Rs Aggarwal 2019 for Class 10 Math Chapter 13 - Trigonometric Identities

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Rs Aggarwal 2019 Solutions for Class 10 Math Chapter 13 Trigonometric Identities are provided here with simple step-by-step explanations. These solutions for Trigonometric Identities are extremely popular among Class 10 students for Math Trigonometric Identities Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Rs Aggarwal 2019 Book of Class 10 Math Chapter 13 are provided here for you for free. You will also love the ad-free experience on Meritnation’s Rs Aggarwal 2019 Solutions. All Rs Aggarwal 2019 Solutions for class Class 10 Math are prepared by experts and are 100% accurate.

Page No 583:

Question 1:

(i) (1 − cos²θ) cosec²θ = 1
(ii) (1 + cot²θ) sin²θ = 1

Answer:

$\begin{array}{l} (i) LHS= (1 - {cos}^{2} θ) {cosec}^{2} θ \\ {=sin}^{2} θ {cosec}^{2} θ (∵ {cos}^{2} θ + {sin}^{2} θ = 1) \\ = \frac{1}{{cosec}^{2} θ} \times {cosec}^{2} θ \\ =1 \\ Hence, LHS = RHS \\ (ii) LHS= (1 + {cot}^{2} θ) {sin}^{2} θ \\ {=cosec}^{2} θ {sin}^{2} θ (∵ {cosec}^{2} θ - {cot}^{2} θ = 1) \\ = \frac{1}{{sin}^{2} θ} \times {sin}^{2} θ \\ =1 \end{array} Hence, LHS = RHS$

Page No 583:

Question 2:

(i) (sec²θ − 1) cot²θ = 1
(ii) (sec²θ − 1) (cosec²θ − 1) = 1
(iii) (1− cos²θ) sec²θ = tan²θ

Answer:

$\begin{array}{l} \begin{array}{l} (i) LHS = (\sec^{2} θ - 1) \cot^{2} θ \\ = \tan^{2} θ \times \cot^{2} θ (∵ \sec^{2} θ - \tan^{2} θ = 1) \\ = \frac{1}{\cot^{2} θ} \times \cot^{2} θ \\ =1 \\ =RH S \end{array} \\ (ii) LHS= (\sec^{2} θ - 1) ({cosec}^{2} θ - 1) \\ {=tan}^{2} θ \times \cot^{2} θ (∵ \sec^{2} θ - \tan^{2} θ = 1 a n d c o \sec^{2} θ - c o t^{2} θ = 1) \\ {=tan}^{2} θ \times \frac{1}{\tan^{2} θ} \\ =1 \\ =RH S \\ (iii) LHS = (1 - \cos^{2} θ) \sec^{2} θ \\ {=sin}^{2} θ \times \sec^{2} θ (∵ \sin^{2} θ + \cos^{2} θ = 1) \\ {=sin}^{2} θ \times \frac{1}{\cos^{2} θ} \\ = \frac{\sin^{2} θ}{\cos^{2} θ} \\ {=tan}^{2} θ \\ =RH S \end{array}$

Page No 583:

Question 3:

(i) $\sin^{2} θ + \frac{1}{(1 + \tan^{2} θ)} = 1$
(ii) $\frac{1}{(1 + \tan^{2} θ)} + \frac{1}{(1 + \cot^{2} θ)} = 1$

Answer:

$\begin{array}{l} (i) {LHS=sin}^{2} θ + \frac{1}{(1 + {tan}^{2} θ)} \\ {=sin}^{2} θ + \frac{1}{{sec}^{2} θ} (∵ {sec}^{2} θ - {tan}^{2} θ = 1) \\ {=sin}^{2} θ + {cos}^{2} θ \\ =1 \\ =RHS \\ (ii) LHS= \frac{1}{(1 + {tan}^{2} θ)} + \frac{1}{({1+cot}^{2} θ)} \\ = \frac{1}{{sec}^{2} θ} + \frac{1}{{cosec}^{2} θ} \\ {=cos}^{2} θ + {sin}^{2} θ \\ =1 \\ =RHS \end{array}$

Page No 583:

Question 4:

(i) (1 + cos θ) (1 − cos θ) (1 + cot²θ) = 1
(ii) cosec θ (1 + cos θ) (cosec θ − cot θ) = 1

Answer:

$\begin{array}{l} (i) LHS= (1 + cos θ) (1 - cos θ) (1 + {cot}^{2} θ) \\ = (1 - {cos}^{2} θ) {cosec}^{2} θ \\ {=sin}^{2} θ \times {cosec}^{2} θ \\ {=sin}^{2} θ \times \frac{1}{{sin}^{2} θ} \\ =1 \\ =RHS \\ (ii) LHS=cosec θ (1 + cos θ) (cosec θ - cot θ) \\ = (cosec θ +cosec θ ×cos θ) (cosec θ - cot θ) \\ = (cosec θ + \frac{1}{sin θ} \times cos θ) (cosec θ - cot θ) \\ = (cosec θ +cot θ) (cosec θ - cot θ) \\ {=cosec}^{2} θ - {cot}^{2} θ (∵ {cosec}^{2} θ - {cot}^{2} θ = 1) \\ =1 \\ =RHS \end{array}$

Page No 583:

Question 5:

Prove each of the following identities:

$(i) \cot^{2} θ - \frac{1}{\sin^{2} θ} = - 1 (ii) \tan^{2} θ - \frac{1}{\cos^{2} θ} = - 1 (iii) \cos^{2} θ + \frac{1}{(1 + \cot^{2} θ)} = 1$

Answer:

$(i) LHS = \cot^{2} θ - \frac{1}{\sin^{2} θ} = \frac{\cos^{2} θ}{\sin^{2} θ} - \frac{1}{\sin^{2} θ} = \frac{\cos^{2} θ - 1}{\sin^{2} θ} = \frac{- \sin^{2} θ}{\sin^{2} θ} = - 1 = RHS$

$(ii) LHS = \tan^{2} θ - \frac{1}{\cos^{2} θ} = \frac{\sin^{2} θ}{\cos^{2} θ} - \frac{1}{\cos^{2} θ} = \frac{\sin^{2} θ - 1}{\cos^{2} θ} = \frac{- \cos^{2} θ}{\cos^{2} θ} = - 1 = RHS$

$(iii) LHS = \cos^{2} θ + \frac{1}{(1 + \cot^{2} θ)} = \cos^{2} θ + \frac{1}{{cosec}^{2} θ} = \cos^{2} θ + \sin^{2} θ = 1 = RHS$

Page No 583:

Question 6:

Prove that $\frac{1}{(1 + \sin θ)} + \frac{1}{(1 - \sin θ)} = 2 \sec^{2} θ$

Answer:

$LHS = \frac{1}{(1 + \sin θ)} + \frac{1}{(1 - \sin θ)} = \frac{(1 - \sin θ) + (1 + \sin θ)}{(1 + \sin θ) (1 - \sin θ)} = \frac{2}{1 - \sin^{2} θ} = \frac{2}{\cos^{2} θ} = 2 \sec^{2} θ = RHS$

Page No 584:

Question 7:

(i) sec θ (1 − sin θ) (sec θ + tan θ) = 1
(ii) sin θ(1 + tan θ) + cos θ(1 + cot θ) = (sec θ + cosec θ)

Answer:

$\begin{array}{l} (i) LHS=sec θ (1 - sin θ) (sec θ + tan θ) \\ = (sec θ - sec θ sin θ) (sec θ + tan θ) \\ = (sec θ - \frac{1}{cos θ} \times sin θ) (sec θ + tan θ) \\ = (sec θ - tan θ) (sec θ + tan θ) \\ {=sec}^{2} θ - {tan}^{2} θ \\ =1 \\ =RHS \\ (ii) LHS=sin θ (1 + tan θ) + cos θ (1 + cot θ) \\ =sin θ + sin θ × \frac{sin θ}{cos θ} + cos θ +cos θ × \frac{cos θ}{sin θ} \\ = \frac{\cos θ {sin}^{2} θ + {sin}^{3} θ + {cos}^{2} θ sin θ {+cos}^{3} θ}{cos θ sin θ} \\ = \frac{({sin}^{3} θ + {cos}^{3} θ) + (cos θ {sin}^{2} θ + {cos}^{2} θ sin θ)}{cos θ sin θ} \\ = \frac{(sin θ + cos θ) ({sin}^{2} θ - sin θ cos θ {+cos}^{2} θ) + sin θ cos θ (sin θ +cos θ)}{cos θ sin θ} \\ = \frac{(sin θ +cos θ) ({sin}^{2} θ + {cos}^{2} θ - sin θ cos θ + sin θ cos θ)}{cos θ sin θ} \\ = \frac{(sin θ +cos θ) (1)}{cos θ sin θ} \\ = \frac{sin θ}{cos θ sin θ} + \frac{cos θ}{cos θ sin θ} \\ = \frac{1}{cos θ} + \frac{1}{sin θ} \\ =sec θ +cosec θ \\ =RHS \end{array}$

Page No 584:

Question 8:

(i) $1 + \frac{\cot^{2} θ}{(1 + cosecθ)} = cosecθ$
(ii) $1 + \frac{\tan^{2} θ}{(1 + secθ)} = secθ$

Answer:

$\begin{array}{l} (i) LHS= 1 + \frac{{cot}^{2} θ}{(1 + cosec θ)} \\ =1+ \frac{({cosec}^{2} θ - 1)}{(cosec θ +1)} (∵ {cosec}^{2} {θ-cot}^{2} θ=1) \\ =1+ \frac{(cosec θ +1) (cosec θ - 1)}{(cosec θ +1)} \\ =1+ (cosec θ - 1) \\ =cosec θ \\ =RHS \\ (ii) LHS=1+ \frac{{tan}^{2} θ}{(1 + sec θ)} \\ =1+ \frac{({sec}^{2} θ - 1)}{(sec θ +1)} \\ =1+ \frac{(sec θ +1) (sec θ - 1)}{(sec θ + 1)} \\ =1+ (sec θ - 1) \\ =sec θ \\ =RHS \end{array}$

Page No 584:

Question 9:

$\frac{(1 + \tan^{2} θ) cotθ}{{cosec}^{2} θ} = \tan θ$

Answer:

$\begin{array}{l} LHS= \frac{(1 + {tan}^{2} θ) cot θ}{{cosec}^{2} θ} \\ = \frac{{sec}^{2} θ cot θ}{{cosec}^{2} θ} \\ = \frac{\frac{1}{{cos}^{2} θ} \times \frac{cos θ}{sin θ}}{\frac{1}{{sin}^{2} θ}} \\ = \frac{1}{cos θ sin θ} \times {sin}^{2} θ \\ = \frac{sin θ}{cos θ} \\ =tan θ \\ =RHS \end{array}$
Hence, L.H.S. = R.H.S.

Page No 584:

Question 10:

$\frac{\tan^{2} θ}{(1 + \tan^{2} θ)} + \frac{\cot^{2} θ}{(1 + \cot^{2} θ)} = 1$

Answer:

$\begin{array}{l} LHS= \frac{{tan}^{2} θ}{(1 + {tan}^{2} θ)} + \frac{{cot}^{2} θ}{(1 + {cot}^{2} θ)} \\ = \frac{{tan}^{2} θ}{{sec}^{2} θ} + \frac{{cot}^{2} θ}{{cosec}^{2} θ} (\begin{array}{l} ∵ {sec}^{2} θ - {tan}^{2} θ = 1 and \\ {cosec}^{2} θ - {cot}^{2} θ = 1 \end{array}) \\ = \frac{\frac{{sin}^{2} θ}{{cos}^{2} θ}}{\frac{1}{{cos}^{2} θ}} + \frac{\frac{{cos}^{2} θ}{{sin}^{2} θ}}{\frac{1}{{sin}^{2} θ}} \\ {=sin}^{2} θ + {cos}^{2} θ \\ =1 \\ =RHS \end{array}$

Hence, LHS = RHS

Page No 584:

Question 11:

$\frac{sinθ}{1 + cosθ} + \frac{(1 + cosθ)}{sinθ} = 2 cosecθ$

Answer:

$\begin{array}{l} LHS= \frac{sin θ}{(1+cos θ)} + \frac{(1+cos θ)}{sin θ} \\ = \frac{{sin}^{2} θ + {(1+cos θ)}^{2}}{(1+cos θ) sin θ} \\ = \frac{{sin}^{2} θ + 1 + {cos}^{2} θ + 2 cos θ}{(1+cos θ) sin θ} \\ = \frac{1 + 1 + 2 cos θ}{(1+cos θ) sin θ} \\ = \frac{2 + 2 cos θ}{(1+cos θ) sin θ} \\ = \frac{2 (1+cos θ)}{(1+cos θ) sin θ} \\ = \frac{2}{sin θ} \\ =2cosec θ \\ =RHS \end{array}$

Hence, LHS = RHS

Page No 584:

Question 12:

$\frac{tanθ}{(1 - cotθ)} + \frac{cotθ}{(1 - tanθ)} = (1 + secθ cosecθ)$

Answer:

$\begin{array}{l} LHS= \frac{tan θ}{(1 - \cot θ)} + \frac{cot θ}{(1 - tan θ)} \\ = \frac{tan θ}{(1 - \frac{cos θ}{sin θ})} + \frac{cot θ}{(1 - \frac{sin θ}{cos θ})} \\ = \frac{sin θ tan θ}{(sin θ - cos θ)} + \frac{cos θ cot θ}{(cos θ - sin θ)} \\ = \frac{sin θ × \frac{sin θ}{cos θ} - cos θ × \frac{cos θ}{sin θ}}{(sin θ - cos θ)} \\ = \frac{\frac{{sin}^{2} θ}{cos θ} - \frac{{cos}^{2} θ}{sin θ}}{(sin θ - cos θ)} \\ = \frac{{sin}^{3} θ - {cos}^{3} θ}{cos θ sin θ (sin θ - cos θ)} \\ = \frac{(sin θ - cos θ) ({sin}^{2} θ + sin θ cos θ {+cos}^{2} θ)}{cos θ sin θ (sin θ - cos θ)} \\ = \frac{1 + sin θ cos θ}{cos θ sin θ} \\ = \frac{1}{cos θ sin θ} + \frac{sin θ cos θ}{cos θ sin θ} \\ =sec θ cosec θ +1 \\ =1+sec θ cosec θ \\ =RHS \end{array}$

Page No 584:

Question 13:

$\frac{\cos^{2} θ}{(1 - tanθ)} + \frac{\sin^{3} θ}{(sinθ - cosθ)} = (1 + sinθ cosθ)$

Answer:

$\frac{\cos^{2} θ}{(1 - tanθ)} + \frac{\sin^{3} θ}{(sinθ - cosθ)} = (1 + sinθ cosθ)$

$\begin{array}{l} LHS= \frac{{cos}^{2} θ}{(1 - tan θ)} + \frac{{sin}^{3} θ}{(sin θ - cos θ)} \\ = \frac{{cos}^{2} θ}{(1 - \frac{sin θ}{cos θ})} + \frac{{sin}^{3} θ}{(sin θ - cos θ)} \\ = \frac{{cos}^{3} θ}{(cos θ - sin θ)} + \frac{{sin}^{3} θ}{(sin θ - cos θ)} \\ = \frac{{cos}^{^{3}} θ - {sin}^{3} θ}{(cos θ - sin θ)} \\ = \frac{(cos θ - sin θ) ({cos}^{2} θ + cos θ sin θ {+sin}^{2} θ)}{(cos θ - sin θ)} \\ = ({sin}^{2} θ + {cos}^{2} θ + cos θ sin θ) \\ = (1 + sin θ cos θ) \\ =RHS \end{array}$

Hence, L.H.S. = R.H.S.

Page No 584:

Question 14:

$\frac{cosθ}{(1 - tanθ)} + \frac{\sin^{2} θ}{(cosθ - sinθ)} = (cosθ + sinθ)$

Answer:

$\begin{array}{l} LHS= \frac{cos θ}{(1 - tan θ)} - \frac{{sin}^{2} θ}{(cos θ - sin θ)} \\ = \frac{cos θ}{(1 - \frac{sin θ}{cos θ})} - \frac{{sin}^{2} θ}{(cos θ - sin θ)} \\ = \frac{{cos}^{2} θ}{(cos θ - sin θ)} - \frac{{sin}^{2} θ}{(cos θ - sin θ)} \\ = \frac{{cos}^{2} θ - {sin}^{2} θ}{(cos θ - sin θ)} \\ = \frac{(cos θ +sin θ) (cos θ - sin θ)}{(cos θ - sin θ)} \\ = (cos θ +sin θ) \\ =RHS \end{array}$

Hence, LHS = RHS

Page No 584:

Question 15:

$(1 + \tan^{2} θ) (1 + \cot^{2} θ) = \frac{1}{(\sin^{2} θ - \sin^{4} θ)}$

Answer:

$\begin{array}{l} LHS= (1 + {tan}^{2} θ) (1 + {cot}^{2} θ) \\ {=sec}^{2} θ . {cosec}^{2} θ (\begin{array}{l} ∵ {sec}^{2} θ - {tan}^{2} θ = 1 and \\ {cosec}^{2} θ - {cot}^{2} θ = 1 \end{array}) \\ = \frac{1}{{cos}^{2} θ . {sin}^{2} θ} \\ = \frac{1}{(1 - {sin}^{2} θ) {sin}^{2} θ} \\ = \frac{1}{{sin}^{2} θ - {sin}^{4} θ} \\ =RHS \end{array}$

Hence, LHS = RHS

Page No 584:

Question 16:

$\frac{tanθ}{(1 + \tan^{2} θ)^{2}} + \frac{cotθ}{(1 + \cot^{2} θ)^{2}} = sinθ cosθ$

Answer:

$\begin{array}{l} LHS= \frac{tan θ}{{(1 + {tan}^{2} θ)}^{2}} + \frac{cot θ}{{(1 + {cot}^{2} θ)}^{2}} \\ = \frac{tan θ}{{({sec}^{2} θ)}^{2}} + \frac{cot θ}{{({cosec}^{2} θ)}^{2}} \\ = \frac{tan θ}{{sec}^{4} θ} + \frac{cot θ}{{cosec}^{4} θ} \\ = \frac{sin θ}{cos θ} \times {cos}^{4} θ + \frac{cos θ}{sin θ} \times {sin}^{4} θ \\ =sin θ {cos}^{3} θ + cos θ {sin}^{3} θ \\ =sin θ cos θ ({cos}^{2} θ + {sin}^{2} θ) \\ =sin θ cos θ \\ =RHS \end{array}$

Hence, LHS = RHS

Page No 584:

Question 17:

(i) $\sin^{6} θ + \cos^{6} θ = 1 - 3 \sin^{2} {θcos}^{2} θ$
(ii) $\sin^{2} θ + \cos^{4} θ = \cos^{2} θ + \sin^{4} θ$
(iii) ${cosec}^{4} θ - {cosec}^{2} θ = \cot^{4} θ + \cot^{2} θ$

Answer:

$\begin{array}{l} (i) {LHS=sin}^{6} θ + {cos}^{6} θ \\ = {({sin}^{2} θ)}^{3} + {({cos}^{2} θ)}^{3} \\ = ({sin}^{2} θ + {cos}^{2} θ) ({sin}^{4} θ - {sin}^{2} θ {cos}^{2} θ + {cos}^{4} θ) \\ =1×{{({sin}^{2} θ)}^{2} + 2 {sin}^{2} θ {cos}^{2} θ + {({cos}^{2} θ)}^{2} - 3 {sin}^{2} θ {cos}^{2} θ} \\ = {({sin}^{2} θ + {cos}^{2} θ)}^{2} - 3 {sin}^{2} θ {cos}^{2} θ \\ = {(1)}^{2} - 3 {sin}^{2} θ {cos}^{2} θ \\ =1 - 3 {sin}^{2} θ {cos}^{2} θ \\ =RHS \end{array} Hence, LHS = RHS \begin{array}{l} (ii) {LHS=sin}^{2} θ + {cos}^{4} θ \\ {=sin}^{2} θ + {({cos}^{2} θ)}^{2} \\ {=sin}^{2} θ + {(1 - {sin}^{2} θ)}^{2} \\ {=sin}^{2} θ + 1 - 2 {sin}^{2} θ + {sin}^{4} θ \\ =1 - {sin}^{2} θ + {sin}^{4} θ \\ {=cos}^{2} θ + {sin}^{4} θ \\ =RHS \end{array} Hence, LHS = RHS \begin{array}{l} (iii) {LHS=cosec}^{4} θ - {cosec}^{2} θ \\ {=cosec}^{2} θ ({cosec}^{2} θ - 1) \\ {=cosec}^{2} θ \times {cot}^{2} θ (∵ {cosec}^{2} θ - {cot}^{2} θ = 1) \\ = (1 + {cot}^{2} θ) {×cot}^{2} θ \\ {=cot}^{2} θ + {cot}^{4} θ \\ =RHS \end{array} Hence, LHS = RHS$

Page No 584:

Question 18:

(i) $\frac{1 - \tan^{2} θ}{1 + \tan^{2} θ} = (\cos^{2} θ - \sin^{2} θ)$
(ii) $\frac{1 - \tan^{2} θ}{\cot^{2} θ - 1} = \tan^{2} θ$

Answer:

$\begin{array}{l} (i) LHS= \frac{1 - {tan}^{2} θ}{1 + {tan}^{2} θ} \\ = \frac{1 - \frac{{sin}^{2} θ}{{cos}^{2} θ}}{1 + \frac{{sin}^{2} θ}{{cos}^{2} θ}} \\ = \frac{{cos}^{2} θ - {sin}^{2} θ}{{cos}^{2} θ + {sin}^{2} θ} \\ = \frac{{cos}^{2} θ - {sin}^{2} θ}{1} \\ {=cos}^{2} θ - {sin}^{2} θ \\ =RHS \\ (ii) LHS= \frac{1 - {tan}^{2} θ}{{cot}^{2} θ - 1} \\ = \frac{1 - \frac{{sin}^{2} θ}{{cos}^{2} θ}}{\frac{{cos}^{2} θ}{{sin}^{2} θ} - 1} \\ = \frac{\frac{{cos}^{2} θ - {sin}^{2} θ}{{cos}^{2} θ}}{\frac{{cos}^{2} θ - {sin}^{2} θ}{{sin}^{2} θ}} \\ = \frac{{sin}^{2} θ}{{cos}^{2} θ} \\ {=tan}^{2} θ \\ =RHS \end{array}$

Page No 584:

Question 19:

(i) $\frac{tanθ}{(secθ - 1)} + \frac{tanθ}{(secθ + 1)} = 2 cosecθ$
(ii) $\frac{cotθ}{(cosecθ + 1)} + \frac{(cosecθ + 1)}{cotθ} = 2 secθ$

Answer:

$\begin{array}{l} (i) LHS= \frac{tan θ}{(sec θ - 1)} + \frac{tan θ}{(sec θ +1)} \\ =tan θ {\frac{sec θ +1+sec θ - 1}{(sec θ - 1) (sec θ +1)}} \\ =tan θ {\frac{2 sec θ}{({sec}^{2} θ - 1)}} \\ =tan θ × \frac{2 sec θ}{{tan}^{2} θ} \\ =2 \frac{sec θ}{tan θ} \\ =2 \frac{\frac{1}{cos θ}}{\frac{sin θ}{cos θ}} \\ =2 \frac{1}{sin θ} \\ =2cosec θ \\ =RHS \end{array} Hence, LHS = RHS \begin{array}{l} (ii) LHS= \frac{cot θ}{(cosec θ +1)} + \frac{(cosec θ + 1)}{cot θ} \\ = \frac{{cot}^{2} θ + {(cosec θ +1)}^{2}}{(cosec θ +1) cot θ} \\ = \frac{{cot}^{2} θ + {cosec}^{2} θ + 2 cosec θ +1}{(cosec θ +1) cot θ} \\ = \frac{{cot}^{2} θ + {cosec}^{2} θ + 2 cosec θ {+cosec}^{2} θ - {cot}^{2} θ}{(cosec θ +1) cot θ} \\ = \frac{2 {cosec}^{2} θ + 2 cosec θ}{(cosec θ +1) cot θ} \\ = \frac{2 cosec θ (cosec θ +1)}{(cosec θ +1) cot θ} \\ = \frac{2 cosec θ}{cot θ} \\ =2× \frac{1}{sin θ} \times \frac{sin θ}{cos θ} \\ =2sec θ \\ =RHS \end{array} Hence, LHS = RHS$

Page No 584:

Question 20:

(i) $\frac{secθ - 1}{secθ + 1} = \frac{\sin^{2} θ}{(1 + cosθ)^{2}}$
(ii) $\frac{secθ - tanθ}{secθ + tanθ} = \frac{\cos^{2} θ}{(1 + sinθ)^{2}}$

Answer:

$\begin{array}{l} (i) LHS= \frac{sec θ - 1}{sec θ +1} \\ = \frac{\frac{1}{cos θ} - 1}{\frac{1}{cos θ} + 1} \\ = \frac{\frac{1 - cos θ}{cos θ}}{\frac{1 + cos θ}{cos θ}} \\ = \frac{1 - cos θ}{1 + cos θ} \\ = \frac{(1 - cos θ) (1 + cos θ)}{(1 + cos θ) (1 + cos θ)} {\begin{cases} Dividing the numerator and \\ denominator by (1 + cos θ) \end{cases}} \\ = \frac{1 - {cos}^{2} θ}{{(1 + cos θ)}^{2}} \\ = \frac{{sin}^{2} θ}{{(1 + cos θ)}^{2}} \\ =RHS \\ (ii) LHS= \frac{sec θ - tan θ}{sec θ +tan θ} \\ = \frac{\frac{1}{cos θ} - \frac{sin θ}{cos θ}}{\frac{1}{cos θ} + \frac{sin θ}{cos θ}} \\ = \frac{\frac{1 - sin θ}{cos θ}}{\frac{1+sin θ}{cos θ}} \\ = \frac{1 - sin θ}{1+sin θ} \\ = \frac{(1 - sin θ) (1+sin θ)}{(1+sin θ) (1+sin θ)} {\begin{array}{l} Dividing the numerator and \\ denominator by (1+sin θ) \end{array}} \\ = \frac{(1 - {sin}^{2} θ)}{{(1+sin θ)}^{2}} \\ = \frac{{cos}^{2} θ}{{(1+sin θ)}^{2}} \\ =RHS \end{array}$

Page No 585:

Question 21:

Prove each of the following identities:

$(i) \sqrt{\frac{1 + \sin θ}{1 - \sin θ}} = (\sec θ + \tan θ) (ii) \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} = (cosec θ - \cot θ) (iii) \sqrt{\frac{1 + \cos θ}{1 - \cos θ}} + \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} = 2 cosec θ$

Answer:

$(i) LHS = \sqrt{\frac{1 + \sin θ}{1 - \sin θ}} = \sqrt{\frac{(1 + \sin θ)}{(1 - \sin θ)} \times \frac{(1 + \sin θ)}{(1 + \sin θ)}} = \sqrt{\frac{{(1 + \sin θ)}^{2}}{1 - \sin^{2} θ}} = \sqrt{\frac{{(1 + \sin θ)}^{2}}{\cos^{2} θ}} = \frac{1 + \sin θ}{\cos θ} = \frac{1}{\cos θ} + \frac{\sin θ}{\cos θ} = (\sec θ + \tan θ) = RHS$

$(ii) LHS = \sqrt{\frac{1 - \cos θ}{1 + \cos θ}} = \sqrt{\frac{(1 - \cos θ)}{(1 + \cos θ)} \times \frac{(1 - \cos θ)}{(1 - \cos θ)}} = \sqrt{\frac{{(1 - \cos θ)}^{2}}{1 - \cos^{2} θ}} = \sqrt{\frac{{(1 - \cos θ)}^{2}}{\sin^{2} θ}} = \frac{1 - \cos θ}{\sin θ} = \frac{1}{\sin θ} - \frac{\cos θ}{\sin θ} = (cosec θ - \cot θ) = RHS$

$\begin{array}{l} (iii) LHS= \sqrt{\frac{1 + cos θ}{1 - cos θ}} + \sqrt{\frac{1 - cos θ}{1 + cos θ}} \\ = \sqrt{\frac{{(1 + cos θ)}^{2}}{(1 - cos θ) (1 + cos θ)}} + \sqrt{\frac{{(1 - cos θ)}^{2}}{(1 + cos θ) (1 - cos θ)}} \\ = \sqrt{\frac{{(1 + cos θ)}^{2}}{(1 - {cos}^{2} θ)}} + \sqrt{\frac{{(1 - cos θ)}^{2}}{(1 - {cos}^{2} θ)}} \\ = \sqrt{\frac{{(1 + cos θ)}^{2}}{{sin}^{2} θ}} + \sqrt{\frac{{(1 - cos θ)}^{2}}{{sin}^{2} θ}} \\ = \frac{(1 + cos θ)}{sin θ} + \frac{(1 - cos θ)}{sin θ} \\ = \frac{1 + cos θ +1 - cos θ}{sin θ} \\ = \frac{2}{sin θ} \\ =2cosec θ \\ =RHS \end{array}$

Page No 585:

Question 22:

$\frac{\cos^{3} θ + \sin^{3} θ}{cosθ + sinθ} + \frac{\cos^{3} θ - \sin^{3} θ}{cosθ - sinθ} = 2$

Answer:

$\begin{array}{l} LHS= \frac{{cos}^{3} θ {+sin}^{3} θ}{cos θ +sin θ} + \frac{{cos}^{3} θ - {sin}^{3} θ}{cos θ - sin θ} \\ = \frac{(cos θ +sin θ) ({cos}^{2} θ - cos θ sin θ {+sin}^{2} θ)}{(cos θ +sin θ)} + \frac{(cos θ - sin θ) ({cos}^{2} θ + cos θ sin θ {+sin}^{2} θ)}{(cos θ - sin θ)} \\ = ({cos}^{2} θ {+sin}^{2} θ - cos θ sin θ) + ({cos}^{2} θ {+sin}^{2} θ + cos θ sin θ) \\ = (1 - cos θ sin θ) + (1 + cos θ sin θ) \\ =2 \\ =RHS \end{array}$
Hence, LHS= RHS

Page No 585:

Question 23:

$\frac{sinθ}{(cotθ + cosecθ)} - \frac{sinθ}{(cotθ - cosecθ)} = 2$

Answer:

$\begin{array}{l} LHS= \frac{sin θ}{(cot θ +cosec θ)} - \frac{sin θ}{(cot θ - cosec θ)} \\ =sin θ {\frac{(cot θ - cosec θ) - (cot θ + cosec θ)}{(cot θ +cosec θ) (cot θ - cosec θ)}} \\ =sin θ {\frac{- 2 cosec θ}{({cot}^{2} θ - {cosec}^{2} θ)}} \\ =sin θ (\frac{- 2 cosec θ}{- 1}) (∵ {cosec}^{2} θ - {cot}^{2} θ = 1) \\ =sin θ .2cosec θ \\ =sin θ ×2× \frac{1}{sin θ} \\ =2 \\ =RHS \end{array}$

Page No 585:

Question 24:

(i) $\frac{sinθ - cosθ}{sinθ + cosθ} + \frac{sinθ + cosθ}{sinθ - cosθ} = \frac{2}{(2 \sin^{2} θ - 1)}$
(ii) $\frac{sinθ + cosθ}{sinθ - cosθ} + \frac{sinθ - cosθ}{sinθ + cosθ} = \frac{2}{(1 - 2 \cos^{2} θ)}$

Answer:

$\begin{array}{l} (i) LHS= \frac{sin θ - cos θ}{sin θ + cos θ} + \frac{sin θ + cos θ}{sin θ - cos θ} \\ = \frac{{(sin θ - cos θ)}^{2} + {(sin θ + cos θ)}^{2}}{(sin θ + cos θ) (sin θ - cos θ)} \\ = \frac{{sin}^{2} θ + {cos}^{2} θ - 2 sin θ cos θ + {sin}^{2} θ + {cos}^{2} θ + 2 sin θ cos θ}{{sin}^{2} θ - {cos}^{2} θ} \\ = \frac{1 + 1}{{sin}^{2} θ - (1 - {sin}^{2} θ)} (∵ {sin}^{2} θ + {cos}^{2} θ = 1) \\ = \frac{2}{{sin}^{2} θ - 1 + {sin}^{2} θ} \\ = \frac{2}{2 {sin}^{2} θ - 1} \\ =RHS \\ (ii) LHS= \frac{sin θ + cos θ}{sin θ - cos θ} + \frac{sin θ - cos θ}{sin θ + cos θ} \\ = \frac{{(sin θ + cos θ)}^{2} + {(sin θ - cos θ)}^{2}}{(sin θ - cos θ) (sin θ + cos θ)} \\ = \frac{{sin}^{2} θ + {cos}^{2} θ + 2 sin θ cos θ + {sin}^{2} θ + {cos}^{2} θ - 2 sin θ cos θ}{({sin}^{2} θ - {cos}^{2} θ)} \\ = \frac{1 + 1}{(1 - {cos}^{2} θ) - {cos}^{2} θ} (∵ {sin}^{2} θ + {cos}^{2} θ = 1) \\ = \frac{2}{1 - 2 {cos}^{2} θ} \\ =RHS \end{array}$

Page No 585:

Question 25:

$\frac{1 + cosθ - \sin^{2} θ}{sinθ (1 + cosθ)} = cotθ$

Answer:

$\begin{array}{l} LHS= \frac{1 + cos θ - {sin}^{2} θ}{sin θ (1 + cos θ)} \\ = \frac{(1 + cos θ) - (1 - {cos}^{2} θ)}{sin θ (1 + cos θ)} \\ = \frac{cos θ {+cos}^{2} θ}{sin θ (1 + cos θ)} \\ = \frac{cos θ (1 + cos θ)}{sin θ (1 + cos θ)} \\ = \frac{cos θ}{sin θ} \\ =cot θ \\ =RHS \end{array}$
Hence, L.H.S. = R.H.S.

Page No 585:

Question 26:

(i) $\frac{cosecθ + cotθ}{cosecθ - cotθ} = (cosecθ + cotθ)^{2} = 1 + 2 \cot^{2} + 2 cosecθ cotθ$
(ii) $\frac{secθ + tanθ}{secθ - tanθ} = (secθ + tanθ)^{2} = 1 + 2 \tan^{2} θ + 2 secθ tanθ$

Answer:

$\begin{array}{l} (i) Here, \frac{cosec θ + cot θ}{cosec θ - cot θ} \\ = \frac{(cosec θ + cot θ) (cosec θ + cot θ)}{(cosec θ - cot θ) (cosec θ + cot θ)} \\ = \frac{{(cosec θ + cot θ)}^{2}}{({cosec}^{2} θ - {cot}^{2} θ)} \\ = \frac{{(cosec θ + cot θ)}^{2}}{1} \\ = {(cosec θ + cot θ)}^{2} \\ Again, {(cosec θ + cot θ)}^{2} \\ {=cosec}^{2} θ + {cot}^{2} θ + 2 cosec θ cot θ \\ {=1+cot}^{2} θ + {cot}^{2} θ + 2 cosec θ cot θ (∵ {cosec}^{2} θ - {cot}^{2} θ = 1) \\ {=1+2cot}^{2} θ + 2 cosec θ cot θ \\ (ii) Here, \frac{sec θ +tan θ}{sec θ - tan θ} \\ = \frac{(sec θ +tan θ) (sec θ +tan θ)}{(sec θ - tan θ) (sec θ +tan θ)} \\ = \frac{{(sec θ +tan θ)}^{2}}{{sec}^{2} θ - {tan}^{2} θ} \\ = \frac{{(sec θ +tan θ)}^{2}}{1} \\ = {(sec θ +tan θ)}^{2} \\ Again, {(sec θ +tan θ)}^{2} \\ {=sec}^{2} θ + {tan}^{2} θ + 2 sec θ tan θ \\ {=1+tan}^{2} θ + {tan}^{2} θ + 2 sec θ tan θ \\ {=1+2tan}^{2} θ + 2 sec θ tan θ \end{array}$

Page No 585:

Question 27:

(i) $\frac{1 + cosθ + sinθ}{1 + cosθ - sinθ} = \frac{1 + sinθ}{cosθ}$
(ii) $\frac{sinθ + 1 - cosθ}{cosθ - 1 + sinθ} = \frac{1 + sinθ}{cosθ}$

Answer:

$\begin{array}{l} (i) LHS= \frac{1 + cos θ +sin θ}{1 + cos θ - sin θ} \\ = \frac{{(1 + cos θ) + sin θ} {(1 + cos θ) + sin θ}}{{(1 + cos θ) - sin θ} {(1 + cos θ) + sin θ}} {Multiplying the numerator and denominator by (1 + cos θ + sinθ)} \\ = \frac{{(1 + cos θ) + sin θ}^{2}}{{{(1 + cos θ)}^{2} - {sin}^{2} θ}} \\ = \frac{1 + {cos}^{2} θ + 2 cos θ {+sin}^{2} θ + 2 sin θ (1 + cos θ)}{1 + {cos}^{2} θ + 2 cos θ - {sin}^{2} θ} \\ = \frac{2 + 2 cos θ + 2 sin θ (1 + cos θ)}{1 + {cos}^{2} θ + 2 cos θ - (1 - {cos}^{2} θ)} \\ = \frac{2 (1 + cos θ) + 2 sin θ (1 + cos θ)}{2 {cos}^{2} θ + 2 cos θ} \\ = \frac{2 (1 + cos θ) (1 + sin θ)}{2 cos θ (1 + cos θ)} \\ = \frac{1 + sin θ}{cos θ} \\ =RHS \\ (ii) LHS= \frac{sin θ +1 - cos θ}{cos θ - 1 + sin θ} \\ = \frac{(sin θ +1 - cos θ) (sin θ +cos θ +1)}{(cos θ - 1 + sin θ) (sin θ +cos θ +1)} {Multiplying and dividing by 1 + cos θ + sinθ } \\ = \frac{{(sin θ + 1)}^{2} - {cos}^{2} θ}{{(sin θ +cos θ)}^{2} - 1^{2}} \\ = \frac{{sin}^{2} θ + 1 + 2 sin θ - {cos}^{2} θ}{{sin}^{2} θ + {cos}^{2} θ + 2 sin θ cos θ - 1} \\ = \frac{{sin}^{2} θ + {sin}^{2} θ + {cos}^{2} θ + 2 sin θ - {cos}^{2} θ}{2 sin θ cos θ} \\ = \frac{2 {sin}^{2} θ + 2 sin θ}{2 sin θ cos θ} \\ = \frac{2 sin θ (1 + sin θ)}{2sin θ cos θ} \\ = \frac{1+sin θ}{cos θ} \\ =RHS \end{array}$

Page No 585:

Question 28:

$\frac{sinθ}{(secθ + tanθ - 1)} + \frac{cosθ}{(cosecθ + cotθ - 1)} = 1$

Answer:

$\begin{array}{l} LHS= \frac{sin θ}{(sec θ +tan θ - 1)} + \frac{cos θ}{(cosec θ +cot θ - 1)} \\ = \frac{sin θ cos θ}{1 + sin θ - cos θ} + \frac{cos θ sin θ}{1 + cos θ - sin θ} \\ =sin θ cos θ [\frac{1}{1 + (sin θ - cos θ)} + \frac{1}{1 - (sin θ - cos θ)}] \\ =sin θ cos θ [\frac{1 - (sin θ - cos θ) + 1 + (sin θ - cos θ)}{{1 + (sin θ - cos θ)} {1 - (sin θ - cos θ)}}] \\ =sin θ cos θ [\frac{1 - sin θ +cos θ +1+sin θ - cos θ}{1 - {(sin θ - cos θ)}^{2}}] \\ = \frac{2 sin θ cos θ}{1 - ({sin}^{2} θ + {cos}^{2} θ - 2 sin θ cos θ)} \\ = \frac{2 sin θ cos θ}{2 sin θ cos θ} \\ =1 \\ =RHS \end{array}$
Hence, LHS = RHS

Page No 585:

Question 29:

$\frac{sinθ + cosθ}{sinθ - cosθ} + \frac{sinθ - cosθ}{sinθ + cosθ} = \frac{2}{(\sin^{2} θ - \cos^{2} θ)} = \frac{2}{(2 \sin^{2} θ - 1)}$

Answer:

$\begin{array}{l} We have \frac{sin θ +cos θ}{sin θ - cos θ} + \frac{sin θ - cos θ}{sin θ +cos θ} \\ = \frac{{(sin θ +cos θ)}^{2} + {(sin θ - cos θ)}^{2}}{(sin θ - cos θ) (sin θ +cos θ)} \\ = \frac{{sin}^{2} θ + {cos}^{2} θ + 2 sin θ cos θ + {sin}^{2} θ {+cos}^{2} θ - 2 sin θ cos θ}{{sin}^{2} θ - {cos}^{2} θ} \\ = \frac{1 + 1}{{sin}^{2} θ - {cos}^{2} θ} \\ = \frac{2}{{sin}^{2} θ - {cos}^{2} θ} \\ Again, \frac{2}{{sin}^{2} θ - {cos}^{2} θ} \\ = \frac{2}{{sin}^{2} θ - (1 - {sin}^{2} θ)} \\ = \frac{2}{2 {sin}^{2} θ - 1} \end{array}$

Page No 585:

Question 30:

$\frac{cosθ cosecθ - sinθ secθ}{cosθ + sinθ} = cosecθ - secθ$

Answer:

$\begin{array}{l} LHS= \frac{cos θ cosec θ - sin θ sec θ}{cos θ +sin θ} \\ = \frac{\frac{cos θ}{sin θ} - \frac{sin θ}{cos θ}}{cos θ +sin θ} \\ = \frac{{cos}^{2} θ - {sin}^{2} θ}{cos θ sin θ (cos θ +sin θ)} \\ = \frac{(cos θ +sin θ) (cos θ - sin θ)}{cos θ sin θ (cos θ +sin θ)} \\ = \frac{(cos θ - sin θ)}{cos θ sin θ} \\ = \frac{1}{sin θ} - \frac{1}{cos θ} \\ =cosec θ - sec θ \\ =RHS \end{array}$
Hence, LHS = RHS

Page No 585:

Question 31:

$(1 + tanθ + cotθ) (sinθ - cosθ) = (\frac{secθ}{{cosec}^{2} θ} - \frac{cosecθ}{\sec^{2} θ})$

Answer:

$\begin{array}{l} LHS= (1 + tan θ +cot θ) (sin θ - cos θ) \\ =sin θ +tan θ sin θ + cot θ sin θ - cos θ - tan θ cos θ - cot θ cos θ \\ =sin θ +tan θ sin θ + \frac{cos θ}{sin θ} \times sin θ - cos θ - \frac{\sin θ}{cos θ} \times cos θ - cot θ cos θ \\ =sin θ +tan θ sin θ +cos θ - cos θ - sin θ - cot θ cos θ \\ =tan θ sin θ - cot θ cos θ \\ = \frac{sin θ}{cos θ} \times \frac{1}{cosec θ} - \frac{cos θ}{sin θ} \times \frac{1}{sec θ} \\ = \frac{1}{cosec θ} \times \frac{1}{cosec θ} \times sec θ - \frac{1}{sec θ} \times \frac{1}{sec θ} \times cosec θ \\ = \frac{sec θ}{{cosec}^{2} θ} - \frac{cosec θ}{{sec}^{2} θ} \\ =RHS \end{array}$

Hence, LHS = RHS

Page No 585:

Question 32:

Prove that $\frac{\cot^{2} θ (\sec θ - 1)}{(1 + \sin θ)} + \frac{\sec^{2} θ (\sin θ - 1)}{(1 + \sec θ)} = 0$

Answer:

$LHS = \frac{\cot^{2} θ (\sec θ - 1)}{(1 + \sin θ)} + \frac{\sec^{2} θ (\sin θ - 1)}{(1 + \sec θ)} = \frac{\frac{\cos^{2} θ}{\sin^{2} θ} (\frac{1}{\cos θ} - 1)}{(1 + \sin θ)} + \frac{\frac{1}{\cos^{2} θ} (\sin θ - 1)}{(1 + \frac{1}{\cos θ})} = \frac{\frac{\cos^{2} θ}{\sin^{2} θ} (\frac{1 - \cos θ}{\cos θ})}{(1 + \sin θ)} + \frac{\frac{(\sin θ - 1)}{\cos^{2} θ}}{(\frac{\cos θ + 1}{\cos θ})} = \frac{\cos^{2} θ (1 - \cos θ)}{\sin^{2} θ \cos θ (1 + \sin θ)} + \frac{(\sin θ - 1) \cos θ}{(\cos θ + 1) \cos^{2} θ} = \frac{\cos θ (1 - \cos θ)}{(1 - \cos^{2} θ) (1 + \sin θ)} + \frac{(\sin θ - 1) \cos θ}{(\cos θ + 1) (1 - \sin^{2} θ)} = \frac{\cos θ (1 - \cos θ)}{(1 - \cos θ) (1 + \cos θ) (1 + \sin θ)} + \frac{- (1 - \sin θ) \cos θ}{(\cos θ + 1) (1 - \sin θ) (1 + \sin θ)} = \frac{\cos θ}{(1 + \cos θ) (1 + \sin θ)} - \frac{\cos θ}{(\cos θ + 1) (1 + \sin θ)} = 0 = RHS$

Page No 585:

Question 33:

$\{\frac{1}{(\sec^{2} θ - \cos^{2} θ)} + \frac{1}{({cosec}^{2} θ - \sin^{2} θ)}\} (\sin^{2} θ \cos^{2} θ) = \frac{1 - \sin^{2} {θcos}^{2} θ}{2 + \sin^{2} θ \cos^{2} θ}$

Answer:

$\begin{array}{l} LHS= {\frac{1}{{sec}^{2} θ - {cos}^{2} θ} + \frac{1}{{cosec}^{2} θ - {sin}^{2} θ}} ({sin}^{2} θ {cos}^{2} θ) \\ = {\frac{{cos}^{2} θ}{1 - {cos}^{4} θ} + \frac{{sin}^{2} θ}{1 - {sin}^{4} θ}} ({sin}^{2} θ {cos}^{2} θ) \\ = {\frac{{cos}^{2} θ}{(1 - {cos}^{2} θ) ({1+cos}^{2} θ)} + \frac{\sin^{2} θ}{(1 - \sin^{2} θ) (1 + \sin^{2} θ)}} ({sin}^{2} θ {cos}^{2} θ) \\ = [\frac{\cot^{2} θ}{1 + \cos^{2} θ} + \frac{\tan^{2} θ}{1 + \sin^{2} θ}] \sin^{2} θ \cos^{2} θ \\ = \frac{\cos^{4} θ}{1 + \cos^{2} θ} + \frac{\sin^{4} θ}{1 + \sin^{2} θ} \\ = \frac{{(\cos^{2} θ)}^{2}}{1 + \cos^{2} θ} + \frac{{(\sin^{2} θ)}^{2}}{1 + \sin^{2} θ} \\ = \frac{(1 - \sin^{2} θ)}{1 + \cos^{2} θ} + \frac{{(1 - \cos^{2} θ)}^{2}}{1 + \sin^{2} θ} \\ = \frac{{(1 - \sin^{2} θ)}^{2} (1 + \sin^{2}) + {(1 - \cos^{2} θ)}^{2} (1 + {cos}^{2} θ)}{(1 + \sin^{2} θ) (1 + {cos}^{2} θ)} \\ = \frac{\cos^{4} θ (1 + \sin^{2} θ) + \sin^{4} θ (1 + {cos}^{2} θ)}{1 + \sin^{2} θ + \cos^{2} θ + \sin^{2} θ \cos^{2} θ} \end{array} = \frac{\cos^{4} θ + \cos^{4} θ \sin^{2} θ + \sin^{4} θ + \sin^{4} θ {cos}^{2} θ}{1 + 1 + \sin^{2} θ \cos^{2} θ} = \frac{\cos^{4} θ + \sin^{4} θ + \sin^{2} θ \cos^{2} θ (\sin^{2} θ + {cos}^{2} θ)}{2 + \sin^{2} θ \cos^{2} θ} = \frac{(\cos^{2} θ)^{2} + (\sin^{2} θ)^{2} + \sin^{2} θ \cos^{2} θ (1)}{2 + \sin^{2} θ \cos^{2} θ} = \frac{(\cos^{2} θ + \sin^{2} θ)^{2} - 2 \sin^{2} θ \cos^{2} θ + \sin^{2} θ \cos^{2} θ (1)}{2 + \sin^{2} θ \cos^{2} θ} \begin{array}{l} = \frac{1^{2} + \cos^{2} θ \sin^{2} θ - 2 \cos^{2} θ \sin^{2} θ}{2 + \sin^{2} θ \cos^{2} θ} \\ = \frac{1 - \cos^{2} θ \sin^{2} θ}{2 + \sin^{2} θ \cos^{2} θ} \end{array} = RHS$

Page No 585:

Question 34:

$\frac{(\sin A - \sin B)}{(\cos A + \cos B)} + \frac{(\cos A - \cos B)}{(\sin A + \sin B)} = 0$

Answer:

$\begin{array}{l} LHS= \frac{(\sin A - \sin B)}{(cos A + cos B)} + \frac{(cos A - cos B)}{(\sin A + \sin B)} \\ = \frac{(\sin A - \sin B) (\sin A + \sin B) + (cos A - cos B) (cos A - cos B)}{(cos A + cos B) (\sin A + \sin B)} \\ = \frac{{sin}^{2} A - {sin}^{2} B + {cos}^{2} A - {cos}^{2} B}{(cos A + cos B) (\sin A + \sin B)} \end{array} \begin{array}{l} = \frac{0}{(cos A + cos B) (\sin A + \sin B)} \\ =0 \\ =RHS \end{array}$

Page No 586:

Question 35:

$\frac{\tan A + \tan B}{\cot A + \cot B} = \tan A \tan B$

Answer:

$\begin{array}{l} LHS= \frac{\tan A + \tan B}{\cot A + \cot B} \\ = \frac{\tan A + \tan B}{\frac{1}{\tan A} + \frac{1}{\tan B}} \\ = \frac{\tan A + \tan B}{\frac{\tan A + \tan B}{\tan A \tan B}} \\ = \frac{\tan A \tan B (\tan A + \tan B)}{(t an A + \tan B)} \\ = \tan A \tan B \\ =RHS \end{array}$

Hence, LHS = RHS

Page No 586:

Question 36:

Show that none of the following is an identity:
(i) cos²θ + cos θ = 1
(ii) sin2θ + sin θ = 2
(iii) tan²θ + sin θ = cos²θ

Answer:

$\begin{array}{l} (i) {cos}^{2} θ + cos θ =1 \\ {LHS=cos}^{2} θ + cos θ \\ =1 - {sin}^{2} θ + cos θ \\ =1 - ({sin}^{2} θ - cos θ) \\ Since LHS \neq RHS, this is not an identity. \\ (ii) {sin}^{2} θ + sin θ =1 \\ {LHS=sin}^{2} θ + sin θ \\ =1 - {cos}^{2} θ + sin θ \\ =1 - ({cos}^{2} θ - sin θ) \\ Since LHS≠RHS, this is not an identity . \\ (iii) {tan}^{2} θ + sin θ {=cos}^{2} θ \\ {LHS=tan}^{2} θ + sin θ \\ = \frac{{sin}^{2} θ}{{cos}^{2} θ} + sin θ \\ = \frac{1 - {cos}^{2} θ}{{cos}^{2} θ} + sin θ \\ {=sec}^{2} θ - 1 + sin θ \\ Since LHS \neq RHS, this is not an identity . \end{array}$

Page No 586:

Question 37:

Prove that $(\sin θ - 2 \sin^{3} θ) = (2 \cos^{3} θ - \cos θ) \tan θ$

Answer:

$RHS = (2 \cos^{3} θ - \cos θ) \tan θ = (2 \cos^{2} θ - 1) \cos θ \times \frac{\sin θ}{\cos θ} = [2 (1 - \sin^{2} θ) - 1] \sin θ = (2 - 2 \sin^{2} θ - 1) \sin θ = (1 - 2 \sin^{2} θ) \sin θ = (\sin θ - 2 \sin^{3} θ) = LHS$

Page No 586:

Question 38:

If 1 + sin²θ = 3 sinθcosθ then prove that tanθ = 1 or $\frac{1}{2}$ .

Answer:

$1 + \sin^{2} (θ) = 3 \sin (θ) \cos (θ) Divide by \cos^{2} (θ) throughout to get : \sec^{2} (θ) + \tan^{2} (θ) = 3 \tan (θ) \Rightarrow (1 + \tan^{2} (θ)) + \tan^{2} (θ) = 3 \tan (θ) \Rightarrow 2 \tan^{2} (θ) - 3 \tan (θ) + 1 = 0, which is quadratic in \tan (θ) . Solving using Quadratic formula, we get : \tan (θ) = \frac{+ 3 \pm \sqrt{{(- 3)}^{2} - 4 (2) (1)}}{2 (2)} = \frac{+ 3 \pm \sqrt{9 - 8}}{4} = \frac{+ 3 \pm 1}{4} \Rightarrow \tan (θ) = + \frac{4}{4}, + \frac{2}{4} = 1 or \frac{1}{2} .$

Page No 594:

Question 1:

If a cos θ + b sin θ = m and a sin θ − b cos θ = n, prove that (m² + n²) = (a² + b²).

Answer:

$\begin{array}{l} We have m^{2} + n^{2} = [{(a \cos θ + b \sin θ)}^{2} + {(a \sin θ - b \cos θ)}^{2}] \\ = (a^{2} \cos^{2} θ + b^{2} \sin^{2} θ + 2 a b \cos θ \sin θ) + (a^{2} \sin^{2} θ + b^{2} \cos^{2} θ - 2 a b \sin θ \cos θ) \\ = a^{2} \cos^{2} θ + b^{2} \sin^{2} θ + a^{2} \sin^{2} θ + b^{2} \cos^{2} θ \\ =(a^{2} \cos^{2} θ + a^{2} \sin^{2} θ) + (b^{2} \cos^{2} θ + b^{2} \sin^{2} θ) \\ = a^{2} (\cos^{2} θ + \sin^{2} θ) + b^{2} (\cos^{2} θ + \sin^{2} θ) \\ = a^{2} + b^{2} [∵ {sin}^{2} + \cos^{2} = 1] \\ {Hence, m}^{2} + n^{2} = a^{2} + b^{2} \end{array}$

Page No 594:

Question 2:

If x = a sec θ + b tan θ and y = a tan θ + b sec θ, prove that (x² − y²) = (a² − b²).

Answer:

$\begin{array}{l} We have x^{2} - y^{2} = [{(a s e c θ + b \tan θ)}^{2} - {(a \tan θ + b s e c θ)}^{2}] \\ = (a^{2} \sec^{2} θ + b^{2} \tan^{2} θ + 2 a b \sec θ \tan θ) - (a^{2} \tan^{2} θ + b^{2} \sec^{2} θ + 2 a b \tan θ \sec θ) \\ = a^{2} \sec^{2} θ + b^{2} \tan^{2} θ - a^{2} \tan^{2} θ - b^{2} \sec^{2} θ \\ =(a^{2} \sec^{2} θ - a^{2} \tan^{2} θ) - (b^{2} \sec^{2} θ - b^{2} \tan^{2} θ) \\ = a^{2} (\sec^{2} θ - \tan^{2} θ) - b^{2} (\sec^{2} θ - \tan^{2} θ) \\ = a^{2} - b^{2} [∵ \sec^{2} θ - \tan^{2} θ = 1] \\ Hence, x^{2} - y^{2} = a^{2} - b^{2} \end{array}$

Page No 594:

Question 3:

$If (\frac{x}{a} sinθ - \frac{y}{b} cosθ) = 1 and (\frac{x}{a} cosθ + \frac{y}{b} sinθ) = 1,$ prove that $(\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}}) = 2 .$

Answer:

$\begin{array}{l} We have (\frac{x}{a} \sin θ - \frac{y}{b} \cos θ) = 1 \\ Squaring both side, we have: \\ {(\frac{x}{a} \sin θ - \frac{y}{b} \cos θ)}^{2} = {(1)}^{2} \\ \Rightarrow (\frac{x^{2}}{a^{2}} \sin^{2} θ + \frac{y^{2}}{b^{2}} \cos^{2} θ - 2 \frac{x}{a} \times \frac{y}{b} \sin θ \cos θ) = 1 ... (i) \\ Again, (\frac{x}{a} \cos θ + \frac{y}{b} \sin θ) = 1 \\ Squaring both side, we get: \\ {(\frac{x}{a} \cos θ + \frac{y}{b} \sin θ)}^{2} = {(1)}^{2} \\ \Rightarrow (\frac{x^{2}}{a^{2}} \cos^{2} θ + \frac{y^{2}}{b^{2}} \sin^{2} θ + 2 \frac{x}{a} \times \frac{y}{b} \sin θ \cos θ) = 1 ... (ii) \\ Now, adding (i) and (ii), we get : \\ (\frac{x^{2}}{a^{2}} \sin^{2} θ + \frac{y^{2}}{b^{2}} \cos^{2} θ - 2 \frac{x}{a} \times \frac{y}{b} \sin θ \cos θ) + (\frac{x^{2}}{a^{2}} \cos^{2} θ + \frac{y^{2}}{b^{2}} \sin^{2} θ + 2 \frac{x}{a} \times \frac{y}{b} \sin θ \cos θ) = 2 \\ \Rightarrow \frac{x^{2}}{a^{2}} \sin^{2} θ + \frac{y^{2}}{b^{2}} \cos^{2} θ + \frac{x^{2}}{a^{2}} \cos^{2} θ + \frac{y^{2}}{b^{2}} \sin^{2} θ = 2 \\ \Rightarrow (\frac{x^{2}}{a^{2}} \sin^{2} θ + \frac{x^{2}}{a^{2}} \cos^{2} θ) + (\frac{y^{2}}{b^{2}} \cos^{2} θ + \frac{y^{2}}{b^{2}} \sin^{2} θ) = 2 \\ \Rightarrow \frac{x^{2}}{a^{2}} (\sin^{2} θ + \cos^{2} θ) + \frac{y^{2}}{b^{2}} (\cos^{2} θ + \sin^{2} θ) = 2 \\ \Rightarrow \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 2 [∵ \sin^{2} θ + \cos^{2} θ = 1] \\ ∴ \frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 2 \end{array}$

Page No 594:

Question 4:

If (sec θ + tan θ) = m and (sec θ − tan θ) = n, show that mn = 1.

Answer:

$\begin{array}{l} We have (\sec θ + \tan θ) = m ... (i) \\ Again, (\sec θ - \tan θ) = n ... (ii) \\ Now, multiplying (i) and (ii), we get : \end{array} \begin{array}{l} (\sec θ + \tan θ) \times (\sec θ - \tan θ) = m n \\ = > \sec^{2} θ - \tan^{2} θ = m n \\ = > 1 = m n [∵ \sec^{2} θ - \tan^{2} θ = 1] \\ ∴ m n = 1 \end{array}$

Page No 594:

Question 5:

If (cosec θ + cot θ) = m and (cosec θ − cot θ) = n, show that mn = 1.

Answer:

$\begin{array}{l} We have (cosec θ + \cot θ) = m ... (i) \\ Again, (cosec θ - \cot θ) = n ... (ii) \\ Now, multiplying (i) and (ii), we get: \end{array} \begin{array}{l} (cosec θ + \cot θ) \times (cosec θ - \cot θ) = m n \\ = > \cos {ec}^{2} θ - \cot^{2} θ = m n \\ = > 1 = m n [∵ \cos {ec}^{2} θ - \cot^{2} θ = 1] \\ ∴ m n = 1 \end{array}$

Page No 594:

Question 6:

If x = a cos³θ and y = b sin³θ, prove that ${(\frac{x}{a})}^{2 / 3} + {(\frac{y}{b})}^{2 / 3} = 1 .$

Answer:

$\begin{array}{l} We have x = a \cos^{3} θ \\ => \frac{x}{a} = \cos^{3} θ ... (i) \\ Again, y = b \sin^{3} θ \\ => \frac{y}{b} = \sin^{3} θ ... (ii) \\ Now, LHS= {(\frac{x}{a})}^{\frac{2}{3}} + {(\frac{y}{b})}^{\frac{2}{3}} \\ = (\cos^{3} θ)^{\frac{2}{3}} + {(\sin^{3} θ)}^{\frac{2}{3}} [From (i) and (ii)] \\ {=cos}^{2} θ + \sin^{2} θ \\ = 1 \\ Hence, LHS = RHS \end{array}$

Page No 594:

Question 7:

If (tan θ + sin θ) = m and (tan θ − sin θ) = n, prove that (m²− n²)² = 16mn.

Answer:

$\begin{array}{l} We have (\tan θ + \sin θ) = m and (\tan θ - \sin θ) = n \\ Now, LHS= {(m^{2} - n^{2})}^{2} \\ = {[{(\tan θ + \sin θ)}^{2} - {(\tan θ - \sin θ)}^{2}]}^{2} \\ = {[(\tan^{2} θ + \sin^{2} θ + 2 \tan θ \sin θ) - (\tan^{2} θ + \sin^{2} θ - 2 \tan θ \sin θ)]}^{2} \\ = {[(\tan^{2} θ + \sin^{2} θ + 2 \tan θ \sin θ - \tan^{2} θ - \sin^{2} θ + 2 \tan θ \sin θ)]}^{2} \\ = {(4 \tan θ \sin θ)}^{2} \\ = 16 \tan^{2} θ \sin^{2} θ \\ = 16 \frac{\sin^{2} θ}{\cos^{2} θ} \sin^{2} θ \\ = 16 \frac{(1 - \cos^{2} θ) \sin^{2} θ}{\cos^{2} θ} \\ = 16 [\tan^{2} θ (1 - \cos^{2} θ)] \\ = 16 (\tan^{2} θ - \tan^{2} θ \cos^{2} θ) \\ = 16 (\tan^{2} θ - \frac{\sin^{2} θ}{\cos^{2} θ} \times \cos^{2} θ) \\ = 16 (\tan^{2} θ - \sin^{2} θ) \\ = 16 (\tan θ + \sin θ) (\tan θ - \sin θ) \\ = 16 m n [(\tan θ + \sin θ) (\tan θ - \sin θ) = m n] \\ ∴ (m^{2} - n^{2}) {(m^{2} - n^{2})}^{2} = 16 m n \end{array}$

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Question 8:

If (cot θ + tan θ) = m and (sec θ − cos θ) = n, prove that (m²n)^2/3 − (mn²)^2/3 = 1.

Answer:

$\begin{array}{l} We have (\cot θ + \tan θ) = m and (\sec θ - cos θ) = n \\ Now, m^{2} n = [{(\cot θ + \tan θ)}^{2} (\sec θ - cos θ)] \\ = [{(\frac{1}{\tan θ} + \tan θ)}^{2} (\frac{1}{\cos θ} - \cos θ)] \\ = \frac{{(1 + \tan^{2} θ)}^{2}}{\tan^{2} θ} \times \frac{(1 - \cos^{2} θ)}{\cos θ} \\ = \frac{\sec^{4} θ}{\tan^{2} θ} \times \frac{\sin^{2} θ}{\cos θ} \\ = \frac{\sec^{4} θ}{\frac{\sin^{2} θ}{\cos^{2} θ}} \times \frac{\sin^{2} θ}{\cos θ} \\ = \frac{\cos^{2} θ \times \sec^{4} θ}{\cos θ} \\ = \cos θ \sec^{4} θ \\ = \frac{1}{\sec θ} \times \sec^{4} θ = \sec^{3} θ \\ ∴ {(m^{2} n)}^{\frac{2}{3}} = {(\sec^{3} θ)}^{\frac{2}{3}} = \sec^{2} θ \end{array}$

$\begin{array}{l} Again, m n^{2} = [(\cot θ + \tan θ) {(\sec θ - cos θ)}^{2}] \\ = [(\frac{1}{\tan θ} + \tan θ) . {(\frac{1}{\cos θ} - \cos θ)}^{2}] \\ = \frac{(1 + \tan^{2} θ)}{\tan θ} \times \frac{{(1 - \cos^{2} θ)}^{2}}{\cos^{2} θ} \\ = \frac{\sec^{2} θ}{\tan θ} \times \frac{\sin^{4} θ}{\cos^{2} θ} \\ = \frac{\sec^{2} θ}{\frac{\sin θ}{\cos θ}} \times \frac{\sin^{4} θ}{\cos^{2} θ} \\ = \frac{\sec^{2} θ \times \sin^{3} θ}{\cos θ} \\ = \frac{1}{\cos^{2} θ} \times \frac{\sec^{3} θ}{\cos θ} = \tan^{3} θ \\ ∴ {(m n^{2})}^{\frac{2}{3}} = {(\tan^{3} θ)}^{\frac{2}{3}} = \tan^{2} θ \\ Now, {(m^{2} n)}^{\frac{2}{3}} - {(m n^{2})}^{\frac{2}{3}} \\ = \sec^{2} θ - \tan^{2} θ = 1 \\ = RHS \\ Hence proved . \end{array}$

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Question 9:

If (cosec θ − sin θ) = a³and (sec θ − cos θ) = b³, prove that $a^{2} b^{2} (a^{2} + b^{2}) = 1 .$

Answer:

$We have (\cos ec θ -sin θ) = a^{3} \begin{array}{l} => a^{3} = (\frac{1}{\sin θ} - \sin θ) \end{array} \begin{array}{l} => a^{3} = \frac{(1 - \sin^{2} θ)}{\sin θ} \end{array} = \frac{\cos^{2} θ}{\sin θ} ∴ a = \frac{\cos^{\frac{2}{3}} θ}{\sin^{\frac{1}{3}} θ} \begin{array}{l} Again, (\sec θ - \cos θ) = b^{3} \end{array} = > b^{3} = (\frac{1}{\cos θ} - \cos θ) = \frac{(1 - \cos^{2} θ)}{\cos θ} = \frac{\sin^{2} θ}{\cos θ} ∴ b = \frac{\sin^{\frac{2}{3}} θ}{\cos^{\frac{1}{3}} θ} \begin{array}{l} Now, LHS= a^{2} b^{2} (a^{2} + b^{2}) \end{array} = a^{4} b^{2} + a^{2} b^{4} = a^{3} (a b^{2}) + (a^{2} b) b^{3} \begin{array}{ll} = \frac{\cos^{2} θ}{\sin θ} \times [\frac{\cos^{\frac{2}{3}} θ}{\sin^{\frac{1}{3}} θ} \times \frac{\sin^{\frac{4}{3}} θ}{\cos^{\frac{2}{3}} θ}] + [\frac{\cos^{\frac{4}{3}} θ}{\sin^{\frac{2}{3}} θ} \times \frac{\sin^{\frac{2}{3}} θ}{\cos^{\frac{1}{3}} θ}] \times \frac{\sin^{2} θ}{\cos θ} \\ = \frac{\cos^{2} θ}{\sin θ} \times \sin θ + \cos θ \times \frac{\sin^{2} θ}{\cos θ} \end{array} {=cos}^{2} θ + \sin^{2} θ = 1 = RHS Hence proved .$

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Question 10:

If (2 sin θ + 3 cos θ) = 2, show that (3 sin θ − 2 cos θ) = ± 3.

Answer:

$Given, (2 \sin θ + 3 \cos θ) = 2 ... (i) \begin{array}{l} We have {(2 \sin θ + 3 \cos θ)}^{2} + {(3 \sin θ - 2 \cos θ)}^{2} \end{array} {=4sin}^{2} θ + 9 \cos^{2} θ + 12 \sin θ \cos θ + 9 \sin^{2} θ + 4 \cos^{2} θ - 12 \sin θ \cos θ =4 (\sin^{2} θ + \cos^{2} θ) + 9 (\sin^{2} θ + \cos^{2} θ) =4+9 =13 i . e ., {(2 \sin θ + 3 \cos θ)}^{2} + {(3 \sin θ - 2 \cos θ)}^{2} = 13 = > 2^{2} + {(3 \sin θ - 2 \cos θ)}^{2} = 13 = > {(3 \sin θ - 2 \cos θ)}^{2} = 13 - 4 = > {(3 \sin θ - 2 \cos θ)}^{2} = 9 = > (3 \sin θ - 2 \cos θ) = \pm 3$

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Question 11:

If $(\sin θ + \cos θ) = \sqrt{2} \cos θ$ , show that $\cot θ = (\sqrt{2} + 1)$ .

Answer:

$We have, (\sin θ + \cos θ) = \sqrt{2} \cos θ Dividing both sides by \sin θ, we get \frac{\sin θ}{\sin θ} + \frac{\cos θ}{\sin θ} = \frac{\sqrt{2} \cos θ}{\sin θ} \Rightarrow 1 + \cot θ = \sqrt{2} \cot θ \Rightarrow \sqrt{2} \cot θ - \cot θ = 1 \Rightarrow (\sqrt{2} - 1) \cot θ = 1$
$\Rightarrow \cot θ = \frac{1}{(\sqrt{2} - 1)} \Rightarrow \cot θ = \frac{1}{(\sqrt{2} - 1)} \times \frac{(\sqrt{2} + 1)}{(\sqrt{2} + 1)} \Rightarrow \cot θ = \frac{(\sqrt{2} + 1)}{2 - 1} \Rightarrow \cot θ = \frac{(\sqrt{2} + 1)}{1} ∴ \cot θ = (\sqrt{2} + 1)$

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Question 12:

If cos θ + sin θ = $\sqrt{2}$ sin θ, show that sin θ − cos θ = $\sqrt{2}$ cos θ.

Answer:

$Given: cos θ +sin θ = \sqrt{2} \sin θ We have {(\sin θ + \cos θ)}^{2} + {(\sin θ - \cos θ)}^{2} = 2 (\sin^{2} θ + \cos^{2} θ) => {(\sqrt{2} \sin θ)}^{2} + {(\sin θ - \cos θ)}^{2} = 2 {=>2sin}^{2} θ + {(\sin θ - \cos θ)}^{2} = 2 => {(\sin θ - \cos θ)}^{2} = 2 - 2 \sin^{2} θ => {(\sin θ - \cos θ)}^{2} = 2 (1 - \sin^{2} θ) => {(\sin θ - \cos θ)}^{2} = 2 \cos^{2} θ => (\sin θ - \cos θ) = \sqrt{2} \cos θ Hence proved .$

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Question 13:

If $\sec θ + \tan θ = p$ , prove that

$(i) \sec θ = \frac{1}{2} (p + \frac{1}{p}) (ii) \tan θ = \frac{1}{2} (p - \frac{1}{p}) (iii) \sin θ = \frac{p^{2} - 1}{p^{2} + 1}$

Answer:

$(i) We have, \sec θ + \tan θ = p . . . . . (1) \Rightarrow \frac{\sec θ + \tan θ}{1} \times \frac{\sec θ - \tan θ}{\sec θ - \tan θ} = p \Rightarrow \frac{\sec^{2} θ - \tan^{2} θ}{\sec θ - \tan θ} = p \Rightarrow \frac{1}{\sec θ - \tan θ} = p \Rightarrow \sec θ - \tan θ = \frac{1}{p} . . . . . (2) Adding (1) and (2), we get 2 \sec θ = p + \frac{1}{p} \Rightarrow \sec θ = \frac{1}{2} (p + \frac{1}{p})$

$(ii) Subtracting (2) from (1), we get 2 \tan θ = (p - \frac{1}{p}) \Rightarrow \tan θ = \frac{1}{2} (p - \frac{1}{p})$

$(iii) Using (i) and (ii), we get \sin θ = \frac{\tan θ}{s e c θ} = \frac{\frac{1}{2} (p - \frac{1}{p})}{\frac{1}{2} (p + \frac{1}{p})} = \frac{(\frac{p^{2} - 1}{p})}{(\frac{p^{2} + 1}{p})} ∴ \sin θ = \frac{p^{2} - 1}{p^{2} + 1}$

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Question 14:

If tan A = n tan B and sin A = m sin B, prove that cos²A = $\frac{(m^{2} - 1)}{n^{2} - 1} .$

Answer:

$\begin{array}{l} We have \tan A = n \tan B \end{array} => \cot B = \frac{n}{\tan A} ... (i) Again, \sin A = m \sin B = > \cos ec B = \frac{m}{\sin A} ... (ii) S quaring (i) and (ii) and subtracting (ii) from (i), we get \frac{m^{2}}{\sin^{2} A} - \frac{n^{2}}{\tan^{2} A} = \cos e c^{2} B - \cot^{2} B = > \frac{m^{2}}{\sin^{2} A} - \frac{n^{2} \cos}{\sin^{2} A} = 1 = > m^{2} - n^{2} \cos^{2} A = \sin^{2} A = > m^{2} - n^{2} \cos^{2} A = 1 - \cos^{2} A = > n^{2} \cos^{2} A - \cos^{2} A = m^{2} - 1 = > \cos^{2} A (n^{2} - 1) = (m^{2} - 1) = > \cos^{2} A = \frac{(m^{2} - 1)}{(n^{2} - 1)} ∴ \cos^{2} A = \frac{(m^{2} - 1)}{(n^{2} - 1)}$

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Question 15:

If $m = (\cos θ - \sin θ) and n = (\cos θ + \sin θ)$ , then show that $\sqrt{\frac{m}{n}} + \sqrt{\frac{n}{m}} = \frac{2}{\sqrt{1 - \tan^{2} θ}}$ .

Answer:

$LHS = \sqrt{\frac{m}{n}} + \sqrt{\frac{n}{m}} = \frac{\sqrt{m}}{\sqrt{n}} + \frac{\sqrt{n}}{\sqrt{m}} = \frac{m + n}{\sqrt{m n}} = \frac{(\cos θ - \sin θ) + (\cos θ + \sin θ)}{\sqrt{(\cos θ - \sin θ) (\cos θ + \sin θ)}}$
$= \frac{2 \cos θ}{\sqrt{\cos^{2} θ - \sin^{2} θ}} = \frac{(\frac{2 \cos θ}{\cos θ})}{(\frac{\sqrt{\cos^{2} θ - \sin^{2} θ}}{\cos θ})} = \frac{2}{\sqrt{\frac{\cos^{2} θ}{\cos^{2} θ} - \frac{\sin^{2} θ}{\cos^{2} θ}}} = \frac{2}{\sqrt{1 - \tan^{2} θ}} = RHS$

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Question 1:

Write the value of $(1 - \sin^{2} θ) \sec^{2} θ$ .

Answer:

$(1 - \sin^{2} θ) \sec^{2} θ = \cos^{2} θ \times \frac{1}{\cos^{2} θ} = 1$

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Question 2:

Write the value of $(1 - \cos^{2} θ) {cosec}^{2} θ$ .

Answer:

$(1 - \cos^{2} θ) {cosec}^{2} θ = \sin^{2} θ \times \frac{1}{\sin^{2} θ} = 1$

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Question 3:

Write the value of $(1 + \tan^{2} θ) \cos^{2} θ$ .

Answer:

$(1 + \tan^{2} θ) \cos^{2} θ = \sec^{2} θ \times \frac{1}{\sec^{2} θ} = 1$

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Question 4:

Write the value of $(1 + \cot^{2} θ) \sin^{2} θ$ .

Answer:

$(1 + \cot^{2} θ) \sin^{2} θ = {cosec}^{2} θ \times \frac{1}{{cosec}^{2} θ} = 1$

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Question 5:

Write the value of $(\sin^{2} θ + \frac{1}{1 + \tan^{2} θ})$ .

Answer:

$(\sin^{2} θ + \frac{1}{1 + \tan^{2} θ}) = (\sin^{2} θ + \frac{1}{\sec^{2} θ}) = (\sin^{2} θ + \cos^{2} θ) = 1$

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Question 6:

Write the value of $(\cot^{2} θ - \frac{1}{\sin^{2} θ})$ .

Answer:

$(\cot^{2} θ - \frac{1}{\sin^{2} θ}) = (\cot^{2} θ - {cosec}^{2} θ) = - 1$

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Question 7:

Write the value of $\sin θ \cos (90 ° - θ) + \cos θ \sin (90 ° - θ)$ .

Answer:

$\sin θ \cos (90 ° - θ) + \cos θ \sin (90 ° - θ) = \sin θ \sin θ + \cos θ \cos θ = \sin^{2} θ + \cos^{2} θ = 1$

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Question 8:

Write the value of ${cosec}^{2} (90 ° - θ) - \tan^{2} θ$ .

Answer:

${cosec}^{2} (90 ° - θ) - \tan^{2} θ = \sec^{2} θ - \tan^{2} θ = 1$

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Question 9:

Write the value of $\sec^{2} θ (1 + \sin θ) (1 - \sin θ)$ . [CBSE 2009]

Answer:

$\sec^{2} θ (1 + \sin θ) (1 - \sin θ) = \sec^{2} θ (1 - \sin^{2} θ) = \frac{1}{\cos^{2} θ} \times \cos^{2} θ = 1$

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Question 10:

Write the value of $\cos {ec}^{2} θ (1 + \cos θ) (1 - \cos θ)$ .

Answer:

$\cos {ec}^{2} θ (1 + \cos θ) (1 - \cos θ) = {cosec}^{2} θ (1 - \cos^{2} θ) = \frac{1}{\sin^{2} θ} \times \sin^{2} θ = 1$

Disclaimer: The question given in the textbook is incorrect. There should be $\cos θ$ instead $\sin θ$ . The solution provided here is of the same.

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Question 11:

Write the value of $\sin^{2} θ \cos^{2} θ (1 + \tan^{2} θ) (1 + \cot^{2} θ)$ .

Answer:

$\sin^{2} θ \cos^{2} θ (1 + \tan^{2} θ) (1 + \cot^{2} θ) = \sin^{2} θ \cos^{2} θ \sec^{2} θ {cosec}^{2} θ = \sin^{2} θ \times \cos^{2} θ \times \frac{1}{\cos^{2} θ} \times \frac{1}{\sin^{2} θ} = 1$

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Question 12:

Write the value of $(1 + \tan^{2} θ) (1 + \sin θ) (1 - \sin θ)$ . [CBSE 2008]

Answer:

$(1 + \tan^{2} θ) (1 + \sin θ) (1 - \sin θ) = \sec^{2} θ (1 - \sin^{2} θ) = \frac{1}{\cos^{2} θ} \times \cos^{2} θ = 1$

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Question 13:

Write the value of $3 \cot^{2} θ - 3 {cosec}^{2} θ$ .

Answer:

$3 \cot^{2} θ - 3 {cosec}^{2} θ = 3 (\cot^{2} θ - {cosec}^{2} θ) = 3 (- 1) = - 3$

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Question 14:

Write the value of $4 \tan^{2} θ - \frac{4}{\cos^{2} θ}$ .

Answer:

$4 \tan^{2} θ - \frac{4}{\cos^{2} θ} = 4 \tan^{2} θ - 4 \sec^{2} θ = 4 (\tan^{2} θ - \sec^{2} θ) = 4 (- 1) = - 4$

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Question 15:

Write the value of $\frac{\tan^{2} θ - \sec^{2} θ}{\cot^{2} θ - {cosec}^{2} θ}$ .

Answer:

$\frac{\tan^{2} θ - \sec^{2} θ}{\cot^{2} θ - {cosec}^{2} θ} = \frac{- 1}{- 1} = 1$

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Question 16:

$If \sin θ = \frac{1}{2}, then write the value of (3 \cot^{2} θ + 3) . [CBSE 2009]$

Answer:

$As, \sin θ = \frac{1}{2} So, cosec θ = \frac{1}{\sin θ} = 2 . . . . . (i)$

Now,

$3 \cot^{2} θ + 3 = 3 (\cot^{2} θ + 1) = 3 {cosec}^{2} θ = 3 {(2)}^{2} [Using (i)] = 3 (4) = 12$

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Question 17:

$If \cos θ = \frac{2}{3}, the write the value of (4 + 4 \tan^{2} θ) .$

Answer:

$4 + 4 \tan^{2} θ = 4 (1 + \tan^{2} θ) = 4 \sec^{2} θ = \frac{4}{\cos^{2} θ} = \frac{4}{{(\frac{2}{3})}^{2}} = \frac{4}{(\frac{4}{9})} = \frac{4 \times 9}{4} = 9$

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Question 18:

$If \cos θ = \frac{7}{25}, then write the value of (\tan θ + \cot θ) . [CBSE 2008]$

Answer:

$As \sin^{2} θ = 1 - \cos^{2} θ = 1 - {(\frac{7}{25})}^{2} = 1 - \frac{49}{625} = \frac{625 - 49}{625} \Rightarrow \sin^{2} θ = \frac{576}{625} \Rightarrow \sin θ = \sqrt{\frac{576}{625}} \Rightarrow \sin θ = \frac{24}{25}$

$Now, \tan θ + \cot θ = \frac{\sin θ}{\cos θ} + \frac{\cos θ}{\sin θ} = \frac{\sin^{2} θ + \cos^{2} θ}{\cos θ \sin θ} = \frac{1}{(\frac{7}{25} \times \frac{24}{25})} = \frac{1}{(\frac{168}{625})} = \frac{625}{168}$

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Question 19:

$If \cos θ = \frac{2}{3}, then write the value of \frac{(\sec θ - 1)}{(\sec θ + 1)} .$

Answer:

$\frac{\sec θ - 1}{\sec θ + 1} = \frac{(\frac{1}{\cos θ} - \frac{1}{1})}{(\frac{1}{\cos θ} + \frac{1}{1})} = \frac{(\frac{1 - \cos θ}{\cos θ})}{(\frac{1 + \cos θ}{\cos θ})} = \frac{1 - \cos θ}{1 + \cos θ} = \frac{(\frac{1}{1} - \frac{2}{3})}{(\frac{1}{1} + \frac{2}{3})} = \frac{(\frac{1}{3})}{(\frac{5}{3})} = \frac{1}{5}$

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Question 20:

$If 5 \tan θ = 4, then write the value of \frac{(\cos θ - \sin θ)}{(\cos θ + \sin θ)} .$

Answer:

$We have, 5 \tan θ = 4 \Rightarrow \tan θ = \frac{4}{5} Now, \frac{(\cos θ - \sin θ)}{(\cos θ + \sin θ)} = \frac{(\frac{\cos θ}{\cos θ} - \frac{\sin θ}{\cos θ})}{(\frac{\cos θ}{\cos θ} + \frac{\sin θ}{\cos θ})} (Dividing numerator and denominator by \cos θ) = \frac{(1 - \tan θ)}{(1 + \tan θ)} = \frac{(\frac{1}{1} - \frac{4}{5})}{(\frac{1}{1} + \frac{4}{5})} = \frac{(\frac{1}{5})}{(\frac{9}{5})} = \frac{1}{9}$

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Question 21:

$If 3 \cot θ = 4, then write the value of \frac{(2 \cos θ + \sin θ)}{(4 \cos θ - \sin θ)} .$

Answer:

$We have, 3 \cot θ = 4 \Rightarrow \cot θ = \frac{4}{3} Now, \frac{(2 \cos θ + \sin θ)}{(4 \cos θ - \sin θ)} = \frac{(\frac{2 \cos θ}{\sin θ} + \frac{\sin θ}{\sin θ})}{(\frac{4 \cos θ}{\sin θ} - \frac{\sin θ}{\sin θ})} (Dividing numerator and denominator by \sin θ) = \frac{(2 \cot θ + 1)}{(4 \cot θ - 1)} = \frac{(2 \times \frac{4}{3} + 1)}{(4 \times \frac{4}{3} - 1)} = \frac{(\frac{8}{3} + \frac{1}{1})}{(\frac{16}{3} - \frac{1}{1})} = \frac{(\frac{8 + 3}{3})}{(\frac{16 - 3}{3})} = \frac{(\frac{11}{3})}{(\frac{13}{3})} = \frac{11}{13}$

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Question 22:

$If \cot θ = \frac{1}{\sqrt{3}}, then write the value of \frac{(1 - \cos^{2} θ)}{(2 - \sin^{2} θ)} .$

Answer:

$We have, \cot θ = \frac{1}{\sqrt{3}} \Rightarrow \cot θ = \cot (\frac{π}{3}) \Rightarrow θ = \frac{π}{3} Now, \frac{(1 - \cos^{2} θ)}{(2 - \sin^{2} θ)} = \frac{1 - \cos^{2} (\frac{π}{3})}{2 - \sin^{2} (\frac{π}{3})} = \frac{1 - {(\frac{1}{2})}^{2}}{2 - {(\frac{\sqrt{3}}{2})}^{2}} = \frac{(\frac{1}{1} - \frac{1}{4})}{(\frac{2}{1} - \frac{3}{4})} = \frac{(\frac{3}{4})}{(\frac{5}{4})} = \frac{3}{5}$

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Question 23:

$If \tan θ = \frac{1}{\sqrt{5}}, then write the value of \frac{({cosec}^{2} θ - \sec^{2} θ)}{({cosec}^{2} θ + \sec^{2} θ)} .$

Answer:

$\frac{({cosec}^{2} θ - \sec^{2} θ)}{({cosec}^{2} θ + \sec^{2} θ)} = \frac{(1 + \cot^{2} θ) - (1 + \tan^{2} θ)}{(1 + \cot^{2} θ) + (1 + \tan^{2} θ)} = \frac{(1 + \frac{1}{\tan^{2} θ}) - (1 + \tan^{2} θ)}{(1 + \frac{1}{\tan^{2} θ}) + (1 + \tan^{2} θ)} = \frac{(1 + \frac{1}{\tan^{2} θ} - 1 - \tan^{2} θ)}{(1 + \frac{1}{\tan^{2} θ} + 1 + \tan^{2} θ)} = \frac{(\frac{1}{\tan^{2} θ} - \tan^{2} θ)}{(\frac{1}{\tan^{2} θ} + \tan^{2} θ + 2)} = \frac{{(\frac{\sqrt{5}}{1})}^{2} - {(\frac{1}{\sqrt{5}})}^{2}}{{(\frac{\sqrt{5}}{1})}^{2} + {(\frac{1}{\sqrt{5}})}^{2} + 2} = \frac{(\frac{5}{1} - \frac{1}{5})}{(\frac{5}{1} + \frac{1}{5} + \frac{2}{1})} = \frac{(\frac{24}{5})}{(\frac{36}{5})} = \frac{24}{36} = \frac{2}{3}$

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Question 24:

$If \cot A = \frac{4}{3} and (A + B) = 90 °, then what is the value of \tan B ?$

Answer:

$We have, \cot A = \frac{4}{3} \Rightarrow \cot (90 ° - B) = \frac{4}{3} (As, A + B = 90 °) ∴ \tan B = \frac{4}{3}$

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Question 25:

$If \cos B = \frac{3}{5} and (A + B) = 90 °, then find the value of \sin A .$

Answer:

$We have, \cos B = \frac{3}{5} \Rightarrow \cos (90 ° - A) = \frac{3}{5} (As, A + B = 90 °) ∴ \sin A = \frac{3}{5}$

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Question 26:

$If \sqrt{3} \sin θ = \cos θ and θ is an acute angle, then find the value of θ .$

Answer:

$We have, \sqrt{3} \sin θ = \cos θ \Rightarrow \frac{\sin θ}{\cos θ} = \frac{1}{\sqrt{3}} \Rightarrow \tan θ = \frac{1}{\sqrt{3}} \Rightarrow \tan θ = \tan 30 ° ∴ θ = 30 °$

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Question 27:

$Write the value of \tan 10 ° \tan 20 ° \tan 70 ° \tan 80 ° .$

Answer:

$\tan 10 ° \tan 20 ° \tan 70 ° \tan 80 ° = \cot (90 ° - 10 °) \cot (90 ° - 20 °) \tan 70 ° \tan 80 ° = \cot 80 ° \cot 70 ° \tan 70 ° \tan 80 ° = \frac{1}{\tan 80 °} \times \frac{1}{\tan 70 °} \times \tan 70 ° \times \tan 80 ° = 1$

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Question 28:

$Write the value of \tan 1 ° \tan 2 ° . . . \tan 89 ° .$

Answer:

$\tan 1 ° \tan 2 ° . . . \tan 89 ° = \tan 1 ° \tan 2 ° \tan 3 ° . . . \tan 45 ° . . . \tan 87 ° \tan 88 ° \tan 89 ° = \tan 1 ° \tan 2 ° \tan 3 ° . . . \tan 45 ° . . . \cot (90 ° - 87 °) \cot (90 ° - 88 °) \cot (90 ° - 89 °) = \tan 1 ° \tan 2 ° \tan 3 ° . . . \tan 45 ° . . . \cot 3 ° \cot 2 ° \cot 1 ° = \tan 1 ° \times \tan 2 ° \times \tan 3 ° \times . . . \times 1 \times . . . \times \frac{1}{\tan 3 °} \times \frac{1}{\tan 2 °} \times \frac{1}{\tan 1 °} = 1$

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Question 29:

$Write the value of \cos 1 ° \cos 2 ° . . . \cos 180 ° .$

Answer:

$\cos 1 ° \cos 2 ° . . . \cos 180 ° = \cos 1 ° \cos 2 ° . . . \cos 90 ° . . . \cos 180 ° = \cos 1 ° \cos 2 ° . . . 0 . . . \cos 180 ° = 0$

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Question 30:

$If \tan A = \frac{5}{12}, then find the value of (\sin A + \cos A) \sec A . [CBSE 2008]$

Answer:

$(\sin A + \cos A) \sec A = (\sin A + \cos A) \frac{1}{\cos A} = \frac{\sin A}{\cos A} + \frac{\cos A}{\cos A} = \tan A + 1 = \frac{5}{12} + \frac{1}{1} = \frac{5 + 12}{12} = \frac{17}{12}$

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Question 31:

$If \sin θ = \cos (θ - 45 °), where θ is acute, then find the value of θ .$

Answer:

$We have, \sin θ = \cos (θ - 45 °) \Rightarrow \cos (90 ° - θ) = \cos (θ - 45 °) Comparing both sides, we get 90 ° - θ = θ - 45 ° \Rightarrow θ + θ = 90 ° + 45 ° \Rightarrow 2 θ = 135 ° \Rightarrow θ = {(\frac{135}{2})}^{°} ∴ θ = 67.5 °$

Page No 597:

Question 32:

$Find the value of \frac{\sin 50 °}{\cos 40 °} + \frac{cosec 40 °}{\sec 50 °} - 4 \cos 50 ° cosec 40 ° .$

Answer:

$\frac{\sin 50 °}{\cos 40 °} + \frac{cosec 40 °}{\sec 50 °} - 4 \cos 50 ° cosec 40 ° = \frac{\cos (90 ° - 50 °)}{\cos 40 °} + \frac{\sec (90 ° - 40 °)}{\sec 50 °} - 4 \sin (90 ° - 50 °) cosec 40 ° = \frac{\cos 40 °}{\cos 40 °} + \frac{\sec 50 °}{\sec 50 °} - 4 \sin 40 ° \times \frac{1}{\sin 40 °} = 1 + 1 - 4 = - 2$

Page No 597:

Question 33:

$Find the value of \sin 48 ° \sec 42 ° + \cos 48 ° cosec 42 ° .$

Answer:

$\sin 48 ° \sec 42 ° + \cos 48 ° cosec 42 ° = \sin 48 ° cosec (90 ° - 42 °) + \cos 48 ° \sec (90 ° - 42 °) = \sin 48 ° cosec 48 ° + \cos 48 ° \sec 48 ° = \sin 48 ° \times \frac{1}{\sin 48 °} + \cos 48 ° \times \frac{1}{\cos 48 °} = 1 + 1 = 2$

Page No 598:

Question 34:

$If x = a \sin θ and y = b \cos θ, then write the value of (b^{2} x^{2} + a^{2} y^{2}) .$

Answer:

$(b^{2} x^{2} + a^{2} y^{2}) = b^{2} {(a \sin θ)}^{2} + a^{2} {(b \cos θ)}^{2} = b^{2} a^{2} \sin^{2} θ + a^{2} b^{2} \cos^{2} θ = a^{2} b^{2} (\sin^{2} θ + \cos^{2} θ) = a^{2} b^{2} (1) = a^{2} b^{2}$

Page No 598:

Question 35:

$If 5 x = \sec θ and \frac{5}{x} = \tan θ, then find the value of 5 (x^{2} - \frac{1}{x^{2}}) . [CBSE 2010]$

Answer:

$5 (x^{2} - \frac{1}{x^{2}}) = \frac{25}{5} (x^{2} - \frac{1}{x^{2}}) = \frac{1}{5} (25 x^{2} - \frac{25}{x^{2}}) = \frac{1}{5} [{(5 x)}^{2} - {(\frac{5}{x})}^{2}] = \frac{1}{5} [{(\sec θ)}^{2} - {(\tan θ)}^{2}] = \frac{1}{5} (\sec^{2} θ - \tan^{2} θ) = \frac{1}{5} (1) = \frac{1}{5}$

Page No 598:

Question 36:

$If cosec θ = 2 x and \cot θ = \frac{2}{x}, then find the value of 2 (x^{2} - \frac{1}{x^{2}}) . [CBSE 2010]$

Answer:

$2 (x^{2} - \frac{1}{x^{2}}) = \frac{4}{2} (x^{2} - \frac{1}{x^{2}}) = \frac{1}{2} (4 x^{2} - \frac{4}{x^{2}}) = \frac{1}{2} [{(2 x)}^{2} - {(\frac{2}{x})}^{2}] = \frac{1}{2} [{(cosec θ)}^{2} - {(\sec θ)}^{2}] = \frac{1}{2} ({cosec}^{2} θ - \sec^{2} θ) = \frac{1}{2} (1) = \frac{1}{2}$

Page No 598:

Question 37:

$If \sec θ + \tan θ = x, then find the value of \sec θ .$

Answer:

$We have, \sec θ + \tan θ = x . . . . . (i) \Rightarrow \frac{\sec θ + \tan θ}{1} \times \frac{\sec θ - \tan θ}{\sec θ - \tan θ} = x \Rightarrow \frac{\sec^{2} θ - \tan^{2} θ}{\sec θ - \tan θ} = x \Rightarrow \frac{1}{\sec θ - \tan θ} = \frac{x}{1} \Rightarrow \sec θ - \tan θ = \frac{1}{x} . . . . . (ii) Adding (i) and (ii), we get 2 \sec θ = x + \frac{1}{x} \Rightarrow 2 \sec θ = \frac{x^{2} + 1}{x} ∴ \sec θ = \frac{x^{2} + 1}{2 x}$

Page No 598:

Question 38:

$Find the value of \frac{\cos 38 ° cosec 52 °}{\tan 18 ° \tan 35 ° \tan 60 ° \tan 72 ° \tan 55 °} .$

Answer:

$\frac{\cos 38 ° cosec 52 °}{\tan 18 ° \tan 35 ° \tan 60 ° \tan 72 ° \tan 55 °} = \frac{\cos 38 ° \sec (90 ° - 52 °)}{\cot (90 ° - 18 °) \cot (90 ° - 35 °) \tan 60 ° \tan 72 ° \tan 55 °} = \frac{\cos 38 ° \sec 38 °}{\cot 72 ° \cot 55 ° \tan 60 ° \tan 72 ° \tan 55 °} = \frac{\cos 38 ° \times \frac{1}{\cos 38 °}}{\frac{1}{\tan 72 °} \times \frac{1}{\tan 55 °} \times \sqrt{3} \times \tan 72 ° \times \tan 55 °} = \frac{1}{\sqrt{3}}$

Page No 598:

Question 39:

$If \sin θ = x, then write the value of \cot θ .$

Answer:

$\cot θ = \frac{\cos θ}{\sin θ} = \frac{\sqrt{1 - \sin^{2} θ}}{\sin θ} = \frac{\sqrt{1 - x^{2}}}{2}$

Page No 598:

Question 40:

$If \sec θ = x, then write the value of \tan θ .$

Answer:

$As, \tan^{2} θ = \sec^{2} θ - 1 So, \tan θ = \sqrt{\sec^{2} θ - 1} = \sqrt{x^{2} - 1}$

Page No 601:

Question 1:

$Choose the correct answer of the following question : \frac{\sec 30 °}{cosec 60 °} = ? (a) \frac{2}{\sqrt{3}} (b) \frac{\sqrt{3}}{2} (c) \sqrt{3} (d) 1$

Answer:

$\frac{\sec 30 °}{cosec 60 °} = \frac{\sec 30 °}{\sec (90 ° - 60 °)} = \frac{\sec 30 °}{\sec 30 °} = 1 Hence, the correct answer is option (d) .$

Page No 601:

Question 2:

$\frac{\tan 35 °}{\cot 55 °} + \frac{\cot 78 °}{\tan 12 °} = ?$
(a) 0
(b) 1
(c) 2
(d) None of these

Answer:

$(c) 2 We have: \frac{\tan 35^{0}}{\cot 55^{0}} + \frac{\cot 78^{0}}{\tan 12^{o}} = \frac{\tan 35^{0}}{\cot (90^{0} - 35^{0})} + \frac{\cot (90^{0} - 12^{0})}{\tan 12^{0}} = \frac{\tan 35^{0}}{\tan 35^{0}} + \frac{\tan 12^{0}}{\tan 12^{0}} [∵ \cot (90^{0} - θ) = \tan θ] = 1 + 1 = 2$

Page No 601:

Question 3:

tan 10° tan 15° tan 75° tan 80° = ?
(a) $\sqrt{3}$
(b) $\frac{1}{\sqrt{3}}$
(c) −1
(d) 1

Answer:

$(d) 1 We have: {tan10}^{0} \tan 15^{0} \tan 75^{0} \tan 80^{0} = \tan 10^{0} \times \tan 15^{0} \times \tan (90^{0} - 15^{0}) \times \tan (90^{0} - 10^{0}) = \tan 10^{0} \times \tan 15^{0} \times \cot 15^{0} \times \cot 10^{0} [∵ \tan (90^{0} - θ) = \cot θ] = 1$

Page No 601:

Question 4:

tan 5° tan 25° tan 30° tan 65° tan 85° = ?
(a) $\sqrt{3}$
(b) $\frac{1}{\sqrt{3}}$
(c) 1
(d) none of these

Answer:

$The correct option is (b) . We have: {tan5}^{0} \tan 25^{0} \tan 30^{0} . \tan 65^{0} t a n 85^{0} = {tan5}^{0} \tan 25^{0} \tan 30^{0} \tan (90^{0} - 25^{0}) t a n (90^{0} - 5^{0}) = \tan 5^{0} \tan 25^{0} \times \frac{1}{\sqrt{3}} \times \cot 25^{0} \cot 5^{0} [∵ \tan (90^{0} - θ) = \cot θ {and tan30}^{0} = \frac{1}{\sqrt{3}}] = \frac{1}{\sqrt{3}}$

Page No 601:

Question 5:

cos 1° cos 2° cos 3° ... cos 180° = ?
(a) −1
(b) 1
(c) 0
(d) $\frac{1}{2}$

Answer:

$(c) 0 \cos 1^{0} \cos 2^{0} \cos 3^{0} . . . \cos 180^{0} = \cos 1^{0} \cos 2^{0} \cos 3^{0} . . . \cos 90^{0} . . . \cos (180)^{0} = 0 [∵ \cos 90 ° = 0]$

Page No 601:

Question 6:

$\frac{2 \sin^{2} 63 ° + 1 + 2 \sin^{2} 27 °}{3 \cos^{2} 17 ° - 2 + 3 \cos^{2} 73 °} = ?$
(a) $\frac{3}{2}$
(b) $\frac{2}{3}$
(c) 2
(d) 3

Answer:

$(d) 3 Given: \frac{2 \sin^{2} 63^{0} + 1 + 2 \sin^{2} 27^{0}}{3 \cos^{2} 17^{0} - 2 + 3 \cos^{2} 73^{0}} = \frac{2 (\sin^{2} 63^{0} + \sin^{2} 27^{0}) + 1}{3 (\cos^{2} 17^{0} + \cos^{2} 73^{0}) - 2} = \frac{2 [\sin^{2} 63^{0} + \sin^{2} (90^{0} - 63^{0})] + 1}{3 [\cos^{2} 17^{0} + \cos^{2} (90^{0} - 17^{0})] - 2} = \frac{2 (\sin^{2} 63^{0} + \cos^{2} 63^{0}) + 1}{3 (\cos^{2} 17^{0} + \sin^{2} 17^{0}) - 2} [∵ \sin (90^{0} - θ) = \cos θ and \cos (90^{0} - θ) = \sin θ] = \frac{2 \times 1 + 1}{3 \times 1 - 2} [∵ \sin^{2} θ + \cos^{2} θ = 1] = \frac{2 + 1}{3 - 2} = \frac{3}{1} = 3$

Page No 601:

Question 7:

sin 47°cos 43° + cos 47°sin 43° = ?
(a) sin 4°
(b) cos 4°
(c) 1
(d) 0

Answer:

$(c) 1 We have: (\sin 47^{0} \cos 43^{0} + \cos 47^{0} \sin 43^{0}) = \sin 47^{0} \cos (90^{0} - 47^{0}) + \cos 47^{0} \sin (90^{0} - 47^{0}) = \sin 47^{0} \sin 47^{0} + \cos 47^{0} \cos 47^{0} [∵ \cos (90^{0} - θ) = \sin θ and \sin (90^{0} - θ) = \cos θ] = \sin^{2} 47^{0} + \cos^{2} 47^{0} = 1$

Page No 602:

Question 8:

sec 70° sin 20° + cos 20° cosec 70° = ?
(a) 0
(b) 1
(c) −1
(d) 2

Answer:

$(d) 2 We have: {sec70}^{0} \sin 20^{0} + \cos 20^{0} \cos {ec70}^{0} = \frac{{sin20}^{0}}{{cos70}^{0}} + \frac{\cos 20^{0}}{{sin70}^{0}} = \frac{\sin 20^{0}}{\cos (90^{0} - 20^{0})} + \frac{{cos20}^{0}}{\sin (90^{0} - 20^{0})} = \frac{\sin 20^{0}}{\sin 20^{0}} + \frac{{cos20}^{0}}{{cos20}^{0}} [\begin{array}{l} ∵ \cos (90^{0} - θ) = \sin θ and \\ \sin (90^{0} - θ) = \cos θ \end{array}] = 1 + 1 = 2$

Page No 602:

Question 9:

If sin 3A = cos (A − 10°), where 3A is an acute angle, then ∠A = ?
(a) 35°
(b) 25°
(c) 20°
(d) 45°

Answer:

$(b) 25° We have: [\sin 3 A=cos (A - 10^{0})] = > \cos (90^{0} - 3 A) = \cos (A - 10^{0}) [∵ \sin θ = \cos (90^{0} - θ)] = > 90^{0} - 3 A = A - 10^{0} = > - 4 A= - 100 = > A= \frac{10 0^{25}}{4_{1}} = > {A=25}^{0}$

Page No 602:

Question 10:

$Choose the correct answer of the following question : If \sec 4 A = cosec (A - 10 °) and 4 A is acute, then ∠ A = ? (a) 20 ° (b) 30 ° (c) 40 ° (d) 50 °$

Answer:

$We have, \sec 4 A = cosec (A - 10 °) \Rightarrow cosec (90 ° - 4 A) = cosec (A - 10 °) Comparing both sides, we get 90 ° - 4 A = A - 10 ° \Rightarrow 4 A + A = 90 ° + 10 ° \Rightarrow 5 A = 100 ° \Rightarrow A = \frac{100 °}{5} ∴ A = 20 ° Hence, the correct answer is option (a) .$

Page No 602:

Question 11:

If A and B are acute angles such that sin A = cos B then (A + B) = ?
(a) 45°
(b) 60°
(c) 90°
(d) 180°

Answer:

Given that A and B are acute angles such that sin A = cos B.
$\Rightarrow \sin (A) = \sin (90 ° - B) \Rightarrow A = 90 ° - B \Rightarrow A + B = 90 ° .$
Hence, the correct answer is option C.

Page No 602:

Question 12:

If cos(α + β) = 0, then sin(α − β) = ?
(a) sin α
(b) cos β
(c) sin 2α
(d) cos 2β

Answer:

(d) cos 2β

$We have: \cos (α + β) = 0 = > \cos (α + β) = \cos 90^{0} = > α + β = 90^{0} = > α = 90^{0} - β ... (i) Now, sin (α - β) = \sin [(90^{0} - β) - β] [U sing (i)] = \sin (90^{0} - 2 β) = \cos 2 β [∵ \sin (90^{0} - θ) = \cos θ]$

Page No 602:

Question 13:

sin (45° + θ) − cos (45° − θ) = ?
(a) 2 sin θ
(b) 2 cos θ
(c) 0
(d) 1

Answer:

$(c) 0 We have: [\sin (45^{0} + θ) - \cos (45^{0} - θ)] = [\sin {90^{0} - (45^{0} - θ)} - \cos (45^{0} - θ)] =[cos (45^{0} - θ) - \cos (45^{0} - θ)] [∵ \sin (90^{0} - θ) = c o s θ] = 0$

Page No 602:

Question 14:

sec²10° – cot²80 = ?

(a) 1

(b) 0

(c) $\frac{3}{2}$

(d) $\frac{1}{2}$

Answer:

$\sec^{2} (10 °) - \cot^{2} (80 °) = \sec^{2} (10 °) - \cot^{2} (90 ° - 10 °) = \sec^{2} (10 °) - \tan^{2} (10 °) = 1 .$
Hence, the correct answer is option A.

Page No 602:

Question 15:

(cosec²57° − tan²33°) = ?
(a) 0
(b) 1
(c) 2
(d) None of these

Answer:

$(b) 1 We have: (\cos {ec}^{2} 57^{0} - \tan^{2} 33^{0}) = [\cos {ec}^{2} (90^{0} - 33^{0}) - \tan^{2} 33^{0}] = (\sec^{2} 33^{0} - \tan^{2} 33^{0}) [∵ \cos ec (90^{0} - θ) = \sec θ] = 1 [∵ \sec^{2} θ - \tan^{2} θ = 1]$

Page No 602:

Question 16:

$\frac{2 \tan^{2} 30 ° \sec^{2} 52 ° \sin^{2} 38 °}{({cosec}^{2} 70 ° - \tan^{2} 20 °)} = ?$
(a) $\frac{3}{2}$
(b) $\frac{2}{3}$
(c) 2
(d) $\frac{1}{2}$

Answer:

(b) $\frac{2}{3}$

$W e have: [\frac{2 \tan^{2} 30^{0} \sec^{2} 52^{0} \sin^{2} 38^{0}}{\cos e c^{2} 70^{0} - \tan^{2} 20^{0}}] = [\frac{2 \times {(\frac{1}{\sqrt{3}})}^{2} \sec^{2} 52^{0} {\sin^{2} (90^{0} - 52^{0})}}{{\cos {ec}^{2} (90^{0} - 20^{0})} - \tan^{2} 20^{0}}] = [\frac{2}{3} \times \frac{\sec^{2} 52^{0} . \cos^{2} 52^{0}}{\sec^{2} 20^{0} - \tan^{2} 20^{0}}] [∵ \sin (90^{0} - θ) = \cos θ and \cos ec (90^{0} - θ) = \sec θ] = \frac{2}{3} \times \frac{1}{1} [∵ \sec^{2} θ - \tan^{2} θ = 1] = \frac{2}{3}$

Page No 602:

Question 17:

$\{\frac{(\sin^{2} 22 ° + \sin^{2} 68 °)}{(\cos^{2} 22 ° + \cos^{2} 68 °)} + \sin^{2} 63 ° + \cos 63 ° \sin 27\} = ?$
(a) 0
(b) 1
(c) 2
(d) 3

Answer:

$(c) 2 We have: [\frac{\sin^{2} 22^{0} + \sin^{2} 68^{0}}{\cos^{2} 22^{0} + \cos^{2} 68} + \sin^{2} 63^{0} + \cos 63^{0} \sin 27^{0}] = [\frac{\sin^{2} 22^{0} + \sin^{2} (90^{0} - 22^{0})}{\cos^{2} (90^{0} - 68^{0}) + \cos^{2} 68^{0}} + \sin^{2} 63^{0} + \cos 63^{0} {\sin (90^{0} - 63^{0})}] = [\frac{\sin^{2} 22^{0} + \cos^{2} 22^{0}}{\sin^{2} 68^{0} + \cos^{2} 68^{0}} + \sin^{2} 63^{0} + \cos 63^{0} \cos 63^{0}] [∵ \sin (90^{0} - θ) = \cos θ and \cos (90^{0} - θ) = \sin θ] = [\frac{1}{1} + \sin^{2} 63^{0} + \cos^{2} 63^{0}] [∵ \sin^{2} θ + \cos^{2} θ = 1] = 1 + 1 = 2$

Page No 602:

Question 18:

$\frac{\cot (90 ° - θ) \cdot \sin (90 ° - θ)}{sinθ} + \frac{\cot 40 °}{\tan 50 °} - (\cos^{2} 20 ° + \cos^{2} 70 °) = ?$
(a) 3
(b) 2
(c) 1
(d) 0

Answer:

$(c) 1 W e have: [\frac{\cot (90^{0} - θ) . \sin (90^{0} - θ)}{\sin θ} + \frac{\cot 40^{0}}{\tan 50^{0}} - (\cos^{2} 20^{0} + \cos^{2} 70^{0})] = [\frac{\tan θ . \cos θ}{\sin θ} + \frac{\cot (90^{0} - 50^{0})}{\tan 50^{0}} - {\cos^{2} (90^{0} - 70^{0}) + \cos^{2} 70^{0}}] [\begin{array}{l} ∵ \cot (90^{0} - θ) = \tan θ and \\ \sin (90^{0} - θ) = \cos θ \end{array}] = [\frac{\frac{\sin θ}{c o s θ} . c o s θ}{\sin θ} + \frac{\tan 50^{0}}{\tan 50^{0}} - (\sin^{2} 70^{0} + \cos^{2} 70^{0})] [∵ cos (90^{0} - θ) = \sin θ] = (\frac{\sin θ}{\sin θ} + 1 - 1) = 1 + 1 - 1 = 1$

Page No 602:

Question 19:

$\frac{\cos 38 ° cosec 52 °}{\tan 18 ° \tan 35 ° \tan 60 ° \tan 72 ° \tan 55 °} = ?$
(a) $\sqrt{3}$
(b) $\frac{1}{3}$
(c) $\frac{1}{\sqrt{3}}$
(d) $\frac{2}{\sqrt{3}}$

Answer:

(c) $\frac{1}{\sqrt{3}}$

$We have: [\frac{\cos 38^{0} \cos ec 52^{0}}{\tan 18^{0} \tan 35^{0} \tan 60^{0} \tan 72^{0} \tan 55^{0}}] = [\frac{\cos 38^{0} \cos ec (90^{0} - 38^{0})}{\tan 18^{0} \tan 35^{0} \times \sqrt{3} \times \tan (90^{0} - 18^{0}) \tan (90^{0} - 35^{0})}] [\begin{array}{l} ∵ \cos ec (90^{0} - θ) = \sec θ and \\ \tan (90^{0} - θ) = \cot θ \end{array}] = [\frac{\cos 38^{0} \sec 38^{0}}{\tan 18^{0} \tan 35^{0} \times \sqrt{3} \times \cot 18^{0} \cot 35^{0}}] = [\frac{\frac{1}{\sec 38^{0}} \times \sec 38^{0}}{\frac{1}{\cot 18^{0}} \frac{1}{\cot 35^{0}} \times \sqrt{3} \cot 18^{0} \cot 35^{0}}] = \frac{1}{\sqrt{3}}$

Page No 602:

Question 20:

If 2 sin 2θ = $\sqrt{3}$ , then θ = ?
(a) 30°
(b) 45°
(c) 60°
(d) 90°

Answer:

(a) 30^o

$2 \sin 2 θ = \sqrt{3} \Rightarrow \sin 2 θ = \frac{\sqrt{3}}{2} = \sin 60^{o} \Rightarrow \sin 2 θ = \sin 60^{o} \Rightarrow 2 θ = 60^{o} \Rightarrow θ = 30^{o}$

Page No 602:

Question 21:

If 2cos 3θ = 1, then θ = ?
(a) 10°
(b) 15°
(c) 20°
(d) 30°

Answer:

(c) 20^o

$2 \cos 3 θ = 1 \Rightarrow \cos 3 θ = \frac{1}{2} \Rightarrow \cos 3 θ = \cos 60^{o} [∵ \cos 60^{o} = \frac{1}{2}] \Rightarrow 3 θ = 60^{o} \Rightarrow θ = \frac{60^{o}}{3} = 20^{o}$

Page No 603:

Question 22:

If $\sqrt{3}$ tan 2θ − 3 = 0, then θ = ?
(a) 15°
(b) 30°
(c) 45°
(d) 60°

Answer:

(b) 30^o

$\sqrt{3} \tan 2 θ - 3 = 0 \Rightarrow \sqrt{3} \tan 2 θ = 3 \Rightarrow \tan 2 θ = \frac{3}{\sqrt{3}} \Rightarrow \tan 2 θ = \sqrt{3} [∵ \tan 60^{o} = \sqrt{3}] \Rightarrow \tan 2 θ = \tan 60^{o} \Rightarrow 2 θ = 60^{o} \Rightarrow θ = 30^{o}$

Page No 603:

Question 23:

If tan x = 3 cot x, then x = ?
(a) 45°
(b) 60°
(c) 30°
(d) 15°

Answer:

(b) 60^o

$\tan x = 3 c o t x \Rightarrow \frac{\tan x}{\cot x} = 3 \Rightarrow \tan^{2} x = 3 [∵ \cot x = \frac{1}{\tan x}] \Rightarrow \tan x = \sqrt{3} = \tan 60^{o} \Rightarrow x = 60^{o}$

Page No 603:

Question 24:

If x tan 45° cos 60° = sin 60° cot 60°, then x = ?
(a) 1
(b) $\frac{1}{2}$
(c) $\frac{1}{\sqrt{2}}$
(d) $\sqrt{3}$

Answer:

(a) 1

$x \tan 45^{o} \cos 60^{o} = \sin 60^{o} \cot 60^{o} \Rightarrow x (1) (\frac{1}{2}) = (\frac{\sqrt{3}}{2}) (\frac{1}{\sqrt{3}}) \Rightarrow x (\frac{1}{2}) = (\frac{1}{2}) \Rightarrow x = 1$

Page No 603:

Question 25:

If (tan²45° − cos²30°) = x sin 45° cos 45°, then x = ?
(a) 2
(b) −2
(c) $\frac{1}{2}$
(d) $\frac{- 1}{2}$

Answer:

(c) $\frac{1}{2}$

$(\tan^{2} 45^{o} - \cos^{2} 30^{o}) = x \sin 45^{o} \cos 45^{o} \Rightarrow x = \frac{(\tan^{2} 45^{o} - \cos^{2} 30^{o})}{\sin 45^{o} \cos 45^{o}} = \frac{[{(1)}^{2} - {(\frac{\sqrt{3}}{2})}^{2}]}{(\frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}})} = \frac{(1 - \frac{3}{4})}{(\frac{1}{2})} = \frac{(\frac{1}{4})}{(\frac{1}{2})} = \frac{1}{4} \times 2 = \frac{1}{2}$

Page No 603:

Question 26:

(sec²60°− 1) = ?
(a) 2
(b) 3
(c) 4
(d) 0

Answer:

(b) 3

sec²60^o − 1 = (2)² − 1 = 4 − 1 = 3

Page No 603:

Question 27:

(cos 0° + sin 30° + sin 45°) (sin 90° + cos 60° − cos 45°) = ?
(a) $\frac{5}{6}$
(b) $\frac{5}{8}$
(c) $\frac{3}{5}$
(d) $\frac{7}{4}$

Answer:

(d) $\frac{7}{4}$

$(\cos 0^{o} + \sin 30^{o} + \sin 45^{o}) (\sin 90^{o} + \cos 60^{o} - \cos 45^{o}) = (1 + \frac{1}{2} + \frac{1}{\sqrt{2}}) (1 + \frac{1}{2} - \frac{1}{\sqrt{2}}) = (\frac{3}{2} + \frac{1}{\sqrt{2}}) (\frac{3}{2} - \frac{1}{\sqrt{2}}) = [{(\frac{3}{2})}^{2} - {(\frac{1}{\sqrt{2}})}^{2}] = (\frac{9}{4}) - (\frac{1}{2}) = (\frac{9 - 2}{4}) = \frac{7}{4}$

Page No 603:

Question 28:

(sin²30° + 4cot²45° − sec²60°) = ?
(a) 0
(b) $\frac{1}{4}$
(c) 4
(d) 1

Answer:

(b) $\frac{1}{4}$

$(\sin^{2} 30^{o} + 4 \cot^{2} 45^{o} - \sec^{2} 60^{o}) = [{(\frac{1}{2})}^{2} + 4 \times {(1)}^{2} - {(2)}^{2}] = (\frac{1}{4} + 4 - 4) = \frac{1}{4}$

Page No 603:

Question 29:

(3cos²60° + 2cot²30° − 5sin²45°) = ?
(a) $\frac{13}{6}$
(b) $\frac{17}{4}$
(c) 1
(d) 4

Answer:

(b) $\frac{17}{4}$

$(3 \cos^{2} 60^{o} + 2 \cot^{2} 30^{o} - 5 \sin^{2} 45^{o}) = [3 \times {(\frac{1}{2})}^{2} + 2 \times {(\sqrt{3})}^{2} - 5 \times {(\frac{1}{\sqrt{2}})}^{2}] = [\frac{3}{4} + 6 - \frac{5}{2}] = \frac{3 + 24 - 10}{4} = \frac{17}{4}$

Page No 603:

Question 30:

$\cos^{2} 30 ° \cos^{2} 45 ° + 4 \sec^{2} 60 ° + \frac{1}{2} \cos^{2} 90 ° - 2 \tan^{2} 60 ° = ?$
(a) $\frac{73}{8}$
(b) $\frac{75}{8}$
(c) $\frac{81}{8}$
(d) $\frac{83}{8}$

Answer:

(d) $\frac{83}{8}$

$(\cos^{2} 30^{o} \cos^{2} 45^{o} + 4 \sec^{2} 60^{o} + \frac{1}{2} \cos^{2} 90^{o} - 2 \tan^{2} 60^{o}) = [{(\frac{\sqrt{3}}{2})}^{2} \times {(\frac{1}{\sqrt{2}})}^{2} + 4 \times {(2)}^{2} + \frac{1}{2} \times {(0)}^{2} - 2 \times {(\sqrt{3})}^{2}] = [(\frac{3}{4} \times \frac{1}{2}) + 16 - 6] = [\frac{3}{8} + 10] = \frac{3 + 80}{8} = \frac{83}{8}$

Page No 603:

Question 31:

If cosec θ = $\sqrt{10}$ , then sec θ = ?
(a) $\frac{3}{\sqrt{10}}$
(b) $\frac{\sqrt{10}}{3}$
(c) $\frac{1}{\sqrt{10}}$
(d) $\frac{2}{\sqrt{10}}$

Answer:

(b) $\frac{\sqrt{10}}{3}$
Let us first draw a right $∆$ ABC right angled at B and $∠ A = θ$ .
Given: cosec $θ$ = $\sqrt{10}$ , but sin $θ$ = $\frac{1}{\cos e c θ}$ = $\frac{1}{\sqrt{10}}$
Also, sin $θ$ = $\frac{Perpendicular}{Hypotenuse}$ = $\frac{B C}{A C}$
So, $\frac{B C}{A C}$ = $\frac{1}{\sqrt{10}}$
Thus, BC = k and AC = $\sqrt{10}$ k

Using Pythagoras theorem in triangle ABC, we have:
AC² = AB² + BC²
⇒ AB² = AC² $-$ BC²
⇒ AB²= ( $\sqrt{10}$ k)² $-$ (k)²
⇒ AB² = 9k²
⇒ AB = 3k
∴ sec $θ$ = $\frac{A C}{A B}$ = $\frac{\sqrt{10} k}{3 k} = \frac{\sqrt{10}}{3}$

Page No 603:

Question 32:

If tan θ = $\frac{8}{15}$ , then cosec θ = ?
(a) $\frac{17}{8}$
(b) $\frac{8}{17}$
(c) $\frac{17}{15}$
(d) $\frac{15}{17}$

Answer:

(a) $\frac{17}{8}$

Let us first draw a right $∆$ ABC right angled at B and $∠ A = θ$ .
Given: tan $θ$ = $\frac{8}{15}$ , but tan $θ$ = $\frac{B C}{A B}$
So, $\frac{B C}{A B}$ = $\frac{8}{15}$
Thus, BC = 8k and AB = 15k

Using Pythagoras theorem in triangle ABC, we have:
AC² = AB² + BC²
⇒ AC² = (15k)²+ (8k)²
⇒ AC² = 289k²
⇒ AC = 17k
∴ cosec $θ$ = $\frac{A C}{B C} = \frac{17 k}{8 k} = \frac{17}{8}$

Page No 603:

Question 33:

If sin θ $\frac{a}{b}$ , then cos θ = ?
(a) $\frac{b}{\sqrt{b^{2} - a^{2}}}$
(b) $\frac{\sqrt{b^{2} - a^{2}}}{b}$
(c) $\frac{a}{\sqrt{b^{2} - a^{2}}}$
(d) $\frac{b}{a}$

Answer:

(b) $\frac{\sqrt{b^{2} - a^{2}}}{b}$
Let us first draw a right $∆$ ABC right angled at B and $∠ A = θ$ .
Given: sin θ = $\frac{a}{b}$ , but sin θ = $\frac{B C}{A C}$
So, $\frac{B C}{A C}$ = $\frac{a}{b}$
Thus, BC = ak and AC = bk

Using Pythagoras theorem in triangle ABC, we have:
AC² = AB² + BC²
⇒ AB² = AC² $-$ BC²
⇒ AB²= (bk)² $-$ (ak)²
⇒ AB² = $(b^{2} - a^{2})$ k²
⇒ AB = ( $\sqrt{b^{2} - a^{2}}$ )k
∴ cos θ = $\frac{AB}{AC}$ = $\frac{\sqrt{b^{2} - a^{2}} k}{b k}$ = $\frac{\sqrt{b^{2} - a^{2}}}{b}$

Page No 603:

Question 34:

If tan θ = $\sqrt{3}$ , then sec θ = ?
(a) $\frac{2}{\sqrt{3}}$
(b) $\frac{\sqrt{3}}{2}$
(c) $\frac{1}{2}$
(d) 2

Answer:

(d) 2

Let us first draw a right $∆$ ABC right angled at B and $∠ A = θ$ .
Given: tan θ = $\sqrt{3}$
But tan θ = $\frac{B C}{A B}$
So, $\frac{B C}{A B}$ = $\frac{\sqrt{3}}{1}$
Thus, BC = $\sqrt{3}$ k and AB = k

Using Pythagoras theorem, we get:
AC² = AB² + BC²
⇒ AC² = ( $\sqrt{3}$ k)² + (k)²
⇒ AC²= 4k²
⇒ AC = 2k
∴ sec θ = $\frac{AC}{AB} = \frac{2 k}{k} = \frac{2}{1}$

Page No 604:

Question 35:

If sec θ = $\frac{25}{7}$ , then sin θ = ?
(a) $\frac{7}{24}$
(b) $\frac{24}{7}$
(c) $\frac{24}{25}$
(d) None of these

Answer:

(c) $\frac{24}{25}$
Let us first draw a right $∆$ ABC right angled at B and $∠ A = θ$ .
Given: sec θ = $\frac{25}{7}$
But cos θ = $\frac{1}{s e c θ}$ = $\frac{A B}{A C}$ = $\frac{7}{25}$
Thus, AC = 25k and AB = 7k

Using Pythagoras theorem, we get:
AC² = AB² + BC²
⇒ BC² = AC² $-$ AB²
⇒ BC² = (25k)² $-$ (7k)²
⇒ BC²= 576k²
⇒ BC = 24k

∴ sin θ = $\frac{B C}{A C} = \frac{24 k}{25 k} = \frac{24}{25}$

Page No 604:

Question 36:

If sinθ = $\frac{1}{2}$ , then cotθ = ?
(a) $\frac{1}{\sqrt{3}}$
(b) $\sqrt{3}$
(c) $\frac{\sqrt{3}}{2}$
(d) 1

Answer:

(b) $\sqrt{3}$

Given: sinθ = $\frac{1}{2}$ , but sinθ = $\frac{BC}{AC}$
So, $\frac{BC}{AC}$ = $\frac{1}{2}$
Thus, BC = k and AC = 2k

Using Pythagoras theorem in triangle ABC, we have:
AC² = AB² + BC²
AB² = AC² $-$ BC²
AB²= (2k)² $-$ (k)²
AB² = 3k²
AB = $\sqrt{3}$ k
So, tanθ = $\frac{BC}{AB}$ = $\frac{k}{\sqrt{3} k}$ = $\frac{1}{\sqrt{3}}$
∴ cotθ = $\frac{1}{\tan θ}$ = $\sqrt{3}$

Page No 604:

Question 37:

If cosθ = $\frac{4}{5}$ , then tanθ = ?
(a) $\frac{3}{4}$
(b) $\frac{4}{3}$
(c) $\frac{3}{5}$
(d) $\frac{5}{3}$

Answer:

(a) $\frac{3}{4}$
Since cosθ = $\frac{4}{5}$ but cosθ = $\frac{A B}{A C}$
So, $\frac{A B}{A C}$ = $\frac{4}{5}$
Thus, AB = 4k and AC = 5k

Using Pythagoras theorem in triangle ABC, we have:
AC² = AB² + BC²
⇒ BC² = AC² $-$ AB²
⇒ BC²= (5k)² $-$ (4k)²
⇒ BC² = 9k²
⇒ BC = 3k
∴ tanθ = $\frac{B C}{A B}$ = $\frac{3}{4}$

Page No 604:

Question 38:

If 3x = cosec θ and $\frac{3}{x}$ = cot θ, than $3 (x^{2} - \frac{1}{x^{2}}) = ?$
(a) $\frac{1}{27}$
(b) $\frac{1}{81}$
(c) $\frac{1}{3}$
(d) $\frac{1}{9}$

Answer:

(c) $\frac{1}{3}$
Given: 3x = cosec θ and $\frac{3}{x}$ = cot θ
Also, we can deduce that x = $\frac{\cos e c θ}{3}$ and $\frac{1}{x} = \frac{c o t θ}{3}$ .
So, substituting the values of x and $\frac{1}{x}$ in the given expression, we get:
3 $(x^{2} - \frac{1}{x^{2}})$ = 3 $({(\frac{\cos e c θ}{3})}^{2} - {(\frac{c o t θ}{3})}^{2})$
= 3 $((\frac{\cos e c^{2} θ}{9}) - (\frac{c o t^{2} θ}{9}))$
= $\frac{3}{9} (\cos e c^{2} θ - c o t^{2} θ)$
= $\frac{1}{3}$ [By using the identity: $(\cos e c^{2} θ - c o t^{2} θ = 1)$ ]

Page No 604:

Question 39:

If 2x = sec A and $\frac{2}{x}$ = tan A, then $2 (x^{2} - \frac{1}{x^{2}}) = ?$ =?
(a) $\frac{1}{2}$
(b) $\frac{1}{4}$
(c) $\frac{1}{8}$
(d) $\frac{1}{16}$

Answer:

(a) $\frac{1}{2}$
Given: 2x = sec A and $\frac{2}{x}$ = tan A
Also, we can deduce that x = $\frac{\sec A}{2}$ and $\frac{1}{x} = \frac{\tan A}{2}$ .
So, substituting the values of x and $\frac{1}{x}$ in the given expression, we get:
2 $(x^{2} - \frac{1}{x^{2}})$ = 2 $({(\frac{\sec A}{2})}^{2} - {(\frac{\tan A}{2})}^{2})$
= 2 $((\frac{\sec^{2} A}{4}) - (\frac{\tan^{2} A}{4}))$
= $\frac{2}{4} (\sec^{2} A - \tan^{2} A)$
= $\frac{1}{2}$ [By using the identity: $(\sec^{2} θ - \tan^{2} θ = 1)$ ]

Page No 604:

Question 40:

If tan θ = $\frac{4}{3}$ , then (sinθ + cosθ) = ?
(a) $\frac{7}{3}$
(b) $\frac{7}{4}$
(c) $\frac{7}{5}$
(d) $\frac{5}{7}$

Answer:

(c) $\frac{7}{5}$
Let us first draw a right $∆$ ABC right angled at B and $∠ A = θ$ .
tan θ = $\frac{4}{3} = \frac{B C}{A B}$
So, AB = 3k and BC = 4k

Using Pythagoras theorem, we get:
AC² = AB² + BC²
⇒ AC² = (3k)² + (4k)²
⇒ AC² = 25k²
⇒ AC= 5k
Thus, sin θ = $\frac{B C}{A C}$ = $\frac{4}{5}$
and cos θ = $\frac{A B}{A C} = \frac{3}{5}$
∴ (sin θ + cos θ) = ( $\frac{4}{5}$ + $\frac{3}{5}$ ) = $\frac{7}{5}$

Page No 604:

Question 41:

If (tan θ + cot θ) = 5, then (tan² θ + cot² θ) = ?

(a) 23
(b) 24
(c) 25
(d) 27

Answer:

(d) 23
We have (tan θ +cot θ) = 5
Squaring both sides, we get:
(tan θ +cot θ)² = 5²
⇒ tan² θ + cot² θ + 2 tan θ cot θ = 25
⇒ tan² θ + cot² θ + 2 = 25 [∵ tan θ = $\frac{1}{\cot θ}$ ]
⇒ tan² θ + cot² θ = 25 − 2 = 23

Page No 604:

Question 42:

If (cos θ + sec θ) = $\frac{5}{2}$ , then (cos² θ + sec² θ) = ?
(a) $\frac{21}{4}$
(b) $\frac{17}{4}$
(c) $\frac{29}{4}$
(d) $\frac{33}{4}$

Answer:

(b) $\frac{17}{4}$
We have (cos θ +sec θ) = $\frac{5}{2}$
Squaring both sides, we get:
(cos θ + sec θ)² = ( $\frac{5}{2}$ )²
$\Rightarrow$ cos² θ + sec² θ + 2 cos θ sec θ = $\frac{25}{4}$
$\Rightarrow$ cos² θ + sec² θ + 2 = $\frac{25}{4}$ [∵ sec θ = $\frac{1}{\cos θ}$ ]
$\Rightarrow$ cos² θ + sec² θ = $\frac{25}{4}$ − 2 = $\frac{17}{4}$

Page No 604:

Question 43:

If tan θ = $\frac{1}{\sqrt{7}}$ , then $\frac{(\cos e c^{2} θ - s e c^{2} θ)}{(\cos e c^{2} θ + s e c^{2} θ)} = ?$ = ?
(a) $\frac{- 2}{3}$
(b) $\frac{- 3}{4}$
(c) $\frac{2}{3}$
(d) $\frac{3}{4}$

Answer:

(d) $\frac{3}{4}$
$= \frac{{cosec}^{2} θ - \sec^{2} θ}{{cosec}^{2} θ + \sec^{2} θ} = \frac{\sin^{2} θ (\frac{1}{\sin^{2} θ} - \frac{1}{\cos^{2} θ})}{\sin^{2} θ (\frac{1}{\sin^{2} θ} + \frac{1}{\cos^{2} θ})} [Multiplying the numerator and denominator by \sin^{2} θ] = \frac{1 - \tan^{2} θ}{1 + \tan^{2} θ}$
= $\frac{1 - \frac{1}{7}}{1 + \frac{1}{7}} = \frac{6}{8} = \frac{3}{4}$

Page No 604:

Question 44:

If 7 tan θ = 4, then $\frac{(7 sinθ - 3 cosθ)}{(7 sinθ + 3 cosθ)}$ = ?
(a) $\frac{1}{7}$
(b) $\frac{5}{7}$
(c) $\frac{3}{7}$
(d) $\frac{5}{14}$

Answer:

(a) $\frac{1}{7}$

7 tan θ = 4

Now, dividing the numerator and denominator of the given expression by cos θ, we get:
$\frac{\frac{1}{\cos θ} (7 \sin θ - 3 \cos θ)}{\frac{1}{\cos θ} (7 \sin θ + 3 \cos θ)}$

= $\frac{7 \tan θ - 3}{7 \tan θ + 3}$

= $\frac{4 - 3}{4 + 3}$ [∵ 7 tan θ = 4]
= $\frac{1}{7}$

Page No 605:

Question 45:

If 3 cot θ = 4, then $\frac{(5 sinθ + 3 cosθ)}{(5 sinθ - 3 cosθ)}$ = ?
(a) $\frac{1}{3}$
(b) 3
(c) $\frac{1}{9}$
(d) 9

Answer:

(d) 9

We have $\frac{(5 \sin θ + 3 \cos θ)}{(5 \sin θ - 3 \cos θ)}$ .

Dividing the numerator and denominator of the given expression by sin θ, we get:
$\frac{\frac{1}{\sin θ} (5 \sin θ + 3 \cos θ)}{\frac{1}{\sin θ} (5 \sin θ - 3 \cos θ)}$

= $\frac{5 + 3 c o t θ}{5 - 3 co t θ}$

= $\frac{5 + 4}{5 - 4}$ = 9 [∵ 3 cot θ = 4]

Page No 605:

Question 46:

If θ = $\frac{a}{b}$ , then $\frac{(a sinθ - b cosθ)}{(a \sin θ + b cosθ)}$ = ?
(a) $\frac{(a^{2} + b^{2})}{(a^{2} - b^{2})}$
(b) $\frac{(a^{2} - b^{2})}{(a^{2} + b^{2})}$
(c) $\frac{a^{2}}{(a^{2} + b^{2})}$
(d) $\frac{b^{2}}{(a^{2} + b^{2})}$

Answer:

(b) $\frac{(a^{2} - b^{2})}{(a^{2} + b^{2})}$
We have tan θ = $\frac{a}{b}$

Now, dividing the numerator and denominator of the given expression by cos θ, we get:
$\frac{a \sin θ - b \cos θ}{a \sin θ + b \cos θ} = \frac{\frac{1}{\cos θ} (a \sin θ - b \cos θ)}{\frac{1}{\cos θ} (a \sin θ + b \cos θ)} = \frac{a \tan θ - b}{a \tan θ + b} = \frac{\frac{a^{2}}{b} - b}{\frac{a^{2}}{b} + b} = \frac{a^{2} - b^{2}}{a^{2} + b^{2}}$

Page No 605:

Question 47:

If sin A + sin² A = 1, then (cos² A + cos⁴ A) = ?
(a) $\frac{1}{2}$
(b) 1
(c) 2
(d) 3

Answer:

(b) 1
$\begin{array}{l} sin A + {sin}^{2} A = 1 \\ =>sin A = 1 - {sin}^{2} A \\ = > sin A = {cos}^{2} A (∵ 1 - {sin}^{2} A) \\ = > {sin}^{2} A = {cos}^{4} A (Squaring both sides) \\ =>1 - {cos}^{2} A = {cos}^{4} A \\ {=> cos}^{4} A + {cos}^{2} A = 1 \end{array}$

Page No 605:

Question 48:

If cos A + cos² A = 1, then (sin² A + sin⁴ A) = ?
(a) 1
(b) 2
(c) 3
(d) 4

Answer:

(a) 1
$\begin{array}{l} cos A + {cos}^{2} A = 1 \\ =>cos A = 1 - {cos}^{2} A \\ =>cos A = {sin}^{2} A (∵ 1 - {cos}^{2} A = {sin}^{2}) \\ {=>cos}^{2} A = {sin}^{4} A (Squaring both sides) \\ =>1 - {sin}^{2} A = {sin}^{4} A \\ {=> sin}^{4} A + {sin}^{2} A = 1 \end{array}$

Page No 605:

Question 49:

$\sqrt{\frac{1 - \sin A}{1 + \sin A}} = ?$
(a) (sec A + tan A)
(b) (sec A − tan A)
(c) sec A tan A
(d) None to these

Answer:

(b) (sec A − tan A)

$\sqrt{\frac{1 - sin A}{1 + sin A}} = \sqrt{\frac{(1 - sin A)}{(1 + sin A)} \times \frac{(1 - sin A)}{(1 - sin A)}} [Multiplying the denominator and numerator by (1 - sin A)] = \frac{(1 - sin A)}{\sqrt{1 - {sin}^{2} A}} = \frac{(1 + sin A)}{\sqrt{{cos}^{2} A}} = \frac{(1 - sin A)}{cos A} = \frac{1}{cos A} - \frac{sin A}{cos A} =sec A - \tan A$

Page No 605:

Question 50:

$\sqrt{\frac{1 + \cos A}{1 - \cos A}} = ?$
(a) cosec A – cot A
(b) cosec A + cot A
(c) cosec A cot A
(d) none of these

Answer:

$\sqrt{\frac{1 + \cos (A)}{1 - \cos (A)}} = \sqrt{\frac{1 + \cos (A)}{1 - \cos (A)} \times \frac{1 + \cos (A)}{1 + \cos (A)}} = \frac{1 + \cos (A)}{\sqrt{1 - \cos^{2} (A)}} = \frac{1 + \cos (A)}{\sqrt{\sin^{2} (A)}} = \frac{1 + \cos (A)}{\sin (A)} = \frac{1}{\sin (A)} + \frac{\cos (A)}{\sin (A)} = cosec (A) + \cot (A) .$
Hence, the correct answer is option B.

Page No 605:

Question 51:

If tan θ = $\frac{a}{b},$ then $\frac{(cosθ + sinθ)}{(cosθ - sinθ)} = ?$
(a) $\frac{a + b}{a - b}$
(b) $\frac{a - b}{a + b}$
(c) $\frac{b + a}{b - a}$
(d) $\frac{b - a}{b + a}$

Answer:

(c) $\frac{b + a}{b - a}$
$Given: tan θ = \frac{a}{b} Now, \frac{(cos θ + sin θ)}{(cos θ - sin θ)} = \frac{(1 + tan θ)}{(1 - tan θ)} [Dividing the numerator and denominator by cos θ] = \frac{(1 + \frac{a}{b})}{(1 - \frac{a}{b})} = \frac{(\frac{b + a}{b})}{(\frac{b - a}{b})} = \frac{(b + a)}{(b - a)}$

Page No 605:

Question 52:

(cosec θ − cot θ)² = ?
(a) $\frac{1 + cosθ}{1 - cosθ}$
(b) $\frac{1 - cosθ}{1 + cosθ}$
(c) $\frac{1 + sinθ}{1 - sinθ}$
(d) $\frac{1 - sinθ}{1 + sinθ}$

Answer:

(b) $\frac{1 - cosθ}{1 + cosθ}$

${(cosec θ - cot θ)}^{2} = {(\frac{1}{sin θ} - \frac{cos θ}{sin θ})}^{2} = {(\frac{1 - cos θ}{sin θ})}^{2} = \frac{{(1 - cos θ)}^{2}}{{sin}^{2} θ} = \frac{{(1 - cos θ)}^{2}}{(1 - {cos}^{2} θ)} = \frac{{(1 - cos θ)}^{2}}{(1 + cos θ) (1 - cos θ)} = \frac{(1 - cos θ)}{(1 + cos θ)}$

Page No 605:

Question 53:

(sec A + tan A) (1 − sin A) = ?
(a) sin A
(b) cos A
(c) sec A
(d) cosec A

Answer:

(b) cos A

$\begin{array}{l} (sec A + \tan A) (1 - sin A) \\ = (\frac{1}{cos A} + \frac{sin A}{cos A}) (1 - sin A) \\ = (\frac{1 + sin A}{cos A}) (1 - sin A) \\ = (\frac{1 - {sin}^{2} A}{cos A}) \\ = (\frac{{cos}^{2} A}{cos A}) \\ = cos A \end{array}$

Page No 610:

Question 1:

$\frac{\cos^{2} 56 ° + \cos^{2} 34 °}{\sin^{2} 56 ° + \sin^{2} 34 °} + 3 \tan^{2} 56 ° \tan^{2} 34 ° = ?$
(a) $3 \frac{1}{2}$
(b) 4
(c) 6
(d) 5

Answer:

(b) 4

$\frac{\cos^{2} 56^{0} + \cos^{2} 34^{0}}{\sin^{2} 56^{0} + \sin^{2} 34^{0}} + 3 {tan}^{2} 56^{0} {tan}^{2} 34^{0} = \frac{{\cos (90^{0} - 34^{0})}^{2} + \cos^{2} 34^{0}}{{\sin (90^{0} - 34^{0})}^{2} + \sin^{2} 34^{0}} + 3 {\tan (90^{0} - 34^{0})}^{2} {tan}^{2} 34^{0} = \frac{\sin^{2} 34^{0} + \cos^{2} 34^{0}}{\cos^{2} 34^{0} + \sin^{2} 34^{0}} + 3 \cot^{2} 34^{0} {tan}^{2} 34^{0} [\begin{array}{l} ∵ \cos (90^{0} - θ) = \sin θ, \sin (90^{0} - θ) \\ = \cos θ and \tan (90^{0} - θ) = \cot θ \end{array}] = \frac{1}{1} + 3 \times 1 [∵ \cot θ = \frac{1}{\tan θ} and \sin^{2} θ + \cos^{2} θ = 1] =4$

Page No 610:

Question 2:

The value of $(\sin^{2} 30 ° \cos^{2} 45 ° + 4 \tan^{2} 30 ° + \frac{1}{2} \sin^{2} 90 ° + \frac{1}{8} \cot^{2} 60 °) = ?$
(a) $\frac{3}{8}$
(b) $\frac{5}{8}$
(c) 6
(d) 2

Answer:

(d) 2
$(\sin^{2} 30^{0} \cos^{2} 45^{0}) + 4 \tan^{2} 30^{0} + \frac{1}{2} \sin^{2} 90^{0} + \frac{1}{8} \cot^{2} 60^{0} = \frac{1}{2^{2}} \times \frac{1}{{(\sqrt{2})}^{2}} + 4 \times \frac{1}{{(\sqrt{3})}^{2}} + \frac{1}{2} \times 1^{2} + \frac{1}{8} \times \frac{1}{{(\sqrt{3})}^{2}} [\begin{array}{l} ∵ \sin 30^{0} = \frac{1}{2} and \cos 45^{0} = \frac{1}{\sqrt{2}} \\ {and tan30}^{0} = \frac{1}{2} and \cot 60^{0} = \frac{1}{\sqrt{3}} \end{array}] = \frac{1}{4} \times \frac{1}{2} + 4 \times \frac{1}{3} + \frac{1}{2} + \frac{1}{24} = \frac{1}{8} + \frac{4}{3} + \frac{1}{2} + \frac{1}{24} = \frac{3 + 32 + 12 + 1}{24} = \frac{48}{24} = 2$

Page No 610:

Question 3:

If cos A + cos² A = 1, then (sin² A + sin4 A) = ?
(a) $\frac{1}{2}$
(b) 2
(c) 1
(d) 4

Answer:

$\begin{array}{l} (c) 1 \\ cosA +^{2} A =1 \end{array} => \cos A {=sin}^{2} A . . . (i) Squaring both sides of (i), we get : \cos^{2} A = \sin^{4} A . . . (ii) A dding (i) and (ii), we get : \sin^{2} A + \sin^{4} A = \cos A + \cos^{2} A {=>sin}^{2} A + \sin^{4} A = 1 [∵ \cos A {+cos}^{2} A =1]$

Page No 611:

Question 4:

If sin $θ = \frac{\sqrt{3}}{2}$ , then (cosec θ + cot θ) = ?
(a) $(2 + \sqrt{3})$
(b) $2 \sqrt{3}$
(c) $\sqrt{2}$
(d) $\sqrt{3}$

Answer:

(d) $\sqrt{3}$
$Given: sin θ = \frac{\sqrt{3}}{2} and \cos ec θ = \frac{2}{\sqrt{3}} \cos {ec}^{2} θ - \cot^{2} θ = 1 => \cot^{2} θ = \cos {ec}^{2} θ - 1 => \cot^{2} θ = \frac{4}{3} - 1 [Given] => \cot θ = \frac{1}{\sqrt{3}} ∴ \cos ec θ + \cot θ = \frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \frac{\sqrt{3} \times \sqrt{3}}{\sqrt{3}} = \sqrt{3}$

Page No 611:

Question 5:

If cot $A = \frac{4}{5}$ , prove that $\frac{(\sin A + \cos A)}{(\sin A - \cos A)} = 9 .$

Answer:

$Given : cotA = \frac{4}{5} Writing \cot A = \frac{\cos A}{\sin A} and sqauring the equation, we get : \frac{\cos^{2} A}{\sin^{2} A} = \frac{16}{25} =>25 \cos^{2} A = 16 \sin^{2} A =>25 \cos^{2} A = 16 - 16 \cos^{2} A => \cos^{2} A = \frac{16}{41} => cosA = \frac{4}{\sqrt{41}} ∴ \sin^{2} A = 1 - \cos^{2} A =1 - \frac{16}{41} Now, \sin A = \sqrt{\frac{25}{41}} =>sin A = \frac{5}{\sqrt{41}} ∴ LHS= \frac{\sin A + cosA}{\sin A - cosA} = \frac{\frac{5}{\sqrt{41}} + \frac{4}{\sqrt{41}}}{\frac{5}{\sqrt{41}} - \frac{4}{\sqrt{41}}} = \frac{9}{1} =9 = RHS$

Page No 611:

Question 6:

If 2x = sec A and $\frac{2}{x}$ = tan A, prove that $(x^{2} - \frac{1}{x^{2}}) = \frac{1}{4} .$

Answer:

$Given: 2 x = secA => x = \frac{secA}{2} . . . (i) and \frac{2}{x} = tanA => \frac{1}{x} = \frac{tanA}{2} . . . (ii) ∴ x + \frac{1}{x} = \frac{secA}{2} + \frac{tanA}{2} [∵ From (i) and (ii)] Also, x - \frac{1}{x} = \frac{secA}{2} - \frac{tanA}{2} ∴ (x + \frac{1}{x}) (x - \frac{1}{x}) = (\frac{secA}{2} + \frac{tanA}{2}) (\frac{secA}{2} - \frac{tanA}{2}) => x^{2} - \frac{1}{x^{2}} = \frac{1}{4} (\sec^{2} A - \tan^{2} A) ∴ x^{2} - \frac{1}{x^{2}} = \frac{1}{4} \times 1 (∵ \sec^{2} A - \tan^{2} A = 1) = \frac{1}{4} Hence proved.$

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Question 7:

If $\sqrt{3}$ tan θ = 3 sin θ, prove that (sin² θ − cos² θ) = $\frac{1}{3} .$

Answer:

$Given: \sqrt{3} tanθ = 3 sinθ => \frac{\sqrt{3}}{cosθ} = 3 [∵ tanθ = \frac{sinθ}{cosθ}] => cosθ = \frac{\sqrt{3}}{3} => \cos^{2} θ = \frac{3}{9} ∴ \sin^{2} θ = 1 - \frac{3}{9} => \sin^{2} θ = \frac{6}{9} ∴ LHS= \sin^{2} θ - \cos^{2} θ = \frac{6}{9} - \frac{3}{9} [∴ \sin^{2} θ = \frac{6}{9}, \cos^{2} θ = \frac{3}{9}] = \frac{3}{9} = \frac{1}{3} =RHS H ence proved.$

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Question 8:

Prove that $\frac{(\sin^{2} 73 ° + \sin^{2} 17 °)}{(\cos^{2} 28 ° + \cos^{2} 62 °)} = 1 .$

Answer:

$\frac{(\sin^{2} 73 ° + \sin^{2} 17 °)}{(\cos^{2} 28 ° + \cos^{2} 62 °)} = 1 .$

$LHS= \frac{\sin^{2} 73^{0} + \sin^{2} 17^{0}}{\cos^{2} 28^{0} + \cos^{2} 62^{0}} = \frac{{[\sin (90^{0} - 17^{0})]}^{2} + \sin^{2} 17^{0}}{{[\cos (90^{0} - 62^{0})]}^{2} + \cos^{2} 62^{0}} = \frac{\cos^{2} 17^{0} + \sin^{2} 17^{0}}{\sin^{2} 62^{0} + \cos^{2} 62^{0}} = \frac{1}{1} [∵ \sin^{2} θ + \cos^{2} θ = 1] =1=RHS$

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Question 9:

If 2 sin 2θ = $\sqrt{3}$ , prove that θ = 30°.

Answer:

$2sin (2 θ) = \sqrt{3} = > sin (2 θ) = \frac{\sqrt{3}}{2} = > sin (2 θ) = \sin (60^{0}) = > 2 θ = 60^{0} = > θ = \frac{60^{0}}{2} = > θ = 30^{0}$

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Question 10:

Prove that $\sqrt{\frac{1 + \cos A}{1 - \cos A}}$ = (cosec A + cot A).

Answer:

$\sqrt{\frac{1 + \cos A}{1 - \cos A}}$ = (cosec A + cot A).

$LHS = \sqrt{\frac{1 + \cos A}{1 - \cos A}} Multiplying the numerator and denominator by (1 + \cos A), we have : \sqrt{\frac{{(1 + \cos A)}^{2}}{(1 - \cos A) (1 + \cos A)}} = \sqrt{\frac{{(1 + \cos A)}^{2}}{1 - \cos^{2} A}} = \frac{1 + \cos A}{\sqrt{\sin^{2} A}} = \frac{1 + \cos A}{\sin A} = \frac{1}{\sin A} + \frac{\cos A}{\sin A} = \cos ec A + \cot A = RHS Hence proved .$

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Question 11:

If cosec θ + cot θ = p, prove that cos θ = $\frac{(p^{2} - 1)}{(p^{2} + 1)} .$

Answer:

$\cos ec θ + \cot θ = p = > \frac{1}{\sin θ} + \frac{\cos θ}{\sin θ} = p = > \frac{1 + \cos θ}{\sin θ} = p Squaring both sides, we get: {(\frac{1 + \cos θ}{\sin θ})}^{2} = p^{2} = > \frac{{(1 + \cos θ)}^{2}}{\sin^{2} θ} = p^{2} = > \frac{{(1 + \cos θ)}^{2}}{1 - \cos^{2} θ} = p^{2} = > \frac{{(1 + \cos θ)}^{2}}{(1 + \cos θ) (1 - \cos θ)} = p^{2} = > \frac{(1 + \cos θ)}{(1 - \cos θ)} = p^{2} = > 1 + \cos θ = p^{2} (1 - \cos θ) = > 1 + \cos θ = p^{2} - p^{2} \cos θ = > \cos θ (1 + p^{2}) = p^{2} - 1 = > \cos θ = \frac{p^{2} - 1}{p^{2} + 1} Hence proved .$

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Question 12:

Prove that (cosec A − cot A)² = $\frac{(1 - \cos A)}{(1 + \cos A)} .$

Answer:

(cosec A − cot A)² = $\frac{(1 - \cos A)}{(1 + \cos A)} .$

$LHS= {(\cos ec A - \cot A)}^{2} = {(\frac{1}{\sin A} - \frac{\cos A}{\sin A})}^{2} = {(\frac{1 - \cos A}{\sin A})}^{2} = \frac{{(1 - \cos A)}^{2}}{\sin^{2} A} = \frac{{(1 - \cos A)}^{2}}{1 - \cos^{2} A} [∵ \sin^{2} θ + \cos^{2} θ = 1] = \frac{(1 - \cos A) (1 - \cos A)}{(1 - \cos A) (1 + \cos A)} = \frac{(1 - \cos A)}{(1 + \cos A)} = RHS Hence proved .$

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Question 13:

If 5cotθ = 3, show that the value of $(\frac{5 \sin θ - 3 \cos θ}{4 \sin θ + 3 \cos θ}) is \frac{16}{29}$ .

Answer:

$Given : 5 \cot (θ) = 3 \Rightarrow \cot (θ) = \frac{3}{5} = \frac{Base (B)}{Perpendicular (P)} Using Pythagoras Theorem, Hypotenuse (H) = \sqrt{B^{2} + P^{2}} = \sqrt{3^{2} + 5^{2}} = \sqrt{34} ∴ \sin (θ) = \frac{P}{H} = \frac{5}{\sqrt{34}}, \cos (θ) = \frac{B}{H} = \frac{3}{\sqrt{34}} Now, \frac{5 \sin (θ) - 3 \cos (θ)}{4 \sin (θ) + 3 \cos (θ)} = \frac{5 \times \frac{5}{\sqrt{34}} - 3 \times \frac{3}{\sqrt{34}}}{4 \times \frac{5}{\sqrt{34}} + 3 \times \frac{3}{\sqrt{34}}} = \frac{\frac{25}{\sqrt{34}} - \frac{9}{\sqrt{34}}}{\frac{20}{\sqrt{34}} + \frac{9}{\sqrt{34}}} = \frac{\frac{16}{\sqrt{34}}}{\frac{29}{\sqrt{34}}} = \frac{16}{\sqrt{34}} \times \frac{\sqrt{34}}{29} = \frac{16}{29}$

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Question 14:

Prove that (sin 32° cos 58° + cos 32° sin 58°) = 1.

Answer:

(sin 32° cos 58° + cos 32° sin 58°) = 1

${LHS=sin32}^{0} \cos 58^{0} + \cos 32^{0} {sin58}^{0} =sin (90^{0} - 58^{0}) \cos 58^{0} + \cos (90^{0} - 58^{0}) {sin58}^{0} = \cos 58^{0} \times \cos 58^{0} + {sin58}^{0} \times {sin58}^{0} [\begin{array}{l} ∵ sin (90^{0} - θ) = \cos θ, \\ \cos (90^{0} - θ) = \cos θ \end{array}] = \cos^{2} 58^{0} + {sin}^{2} 58^{0} =1 [∵ {sin}^{2} θ + \cos^{2} θ = 1] =RHS$

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Question 15:

If x = a sinθ + bcosθ and y = acosθ – bsinθ, prove that $x^{2} + y^{2} = a^{2} + b^{2} .$

Answer:

$Given: x = a \sin θ + b \cos θ Squaring both sides, we get: x^{2} = a^{2} \sin^{2} θ + 2 a b \sin θ \cos θ + b^{2} \cos^{2} θ . . . (i) Also, y = a \cos θ - b \sin θ Squaring both sides, we get: y^{2} = a^{2} \cos^{2} θ - 2 a b \sin θ \cos θ + b^{2} \sin^{2} θ . . . (ii) ∴ LHS= x^{2} + y^{2} = a^{2} \sin^{2} θ + 2 a b \sin θ \cos θ + b^{2} \cos^{2} θ + a^{2} \cos^{2} θ - 2 a b \sin θ \cos θ + b^{2} \sin^{2} θ [using (i) and (ii)] = a^{2} (\sin^{2} θ + \cos^{2} θ) + b^{2} (\sin^{2} θ + \cos^{2} θ) = a^{2} + b^{2} [∵ \sin^{2} θ + \cos^{2} θ = 1] =RHS Hence proved.$

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Question 16:

Prove that $\frac{(1 + sinθ)}{(1 - sinθ)}$ = (sec θ + tan θ)².

Answer:

$\frac{(1 + sinθ)}{(1 - sinθ)}$ = (sec θ + tan θ)²

$LHS= \frac{(1 + \sin θ)}{(1 - \sin θ)} Multiplying the numerator and denominator by (1 + \sin θ), we get : \frac{{(1 + \sin θ)}^{2}}{1 - \sin^{2} θ} = \frac{1 + 2 \sin θ + \sin^{2} θ}{\cos^{2} θ} [∵ \sin^{2} θ + \cos^{2} θ = 1] {=sec}^{2} θ + 2 \times \frac{\sin θ}{\cos θ} \times sec θ + \tan^{2} θ {=sec}^{2} θ + 2 \times \tan θ \times sec θ + \tan^{2} θ = {(sec θ +tan θ)}^{2} =RHS Hence proved .$

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Question 17:

Prove that $\frac{1}{(secθ - tanθ)} - \frac{1}{cosθ} = \frac{1}{cosθ} - \frac{1}{(secθ + tanθ)}$ .

Answer:

$\frac{1}{(secθ - tanθ)} - \frac{1}{cosθ} = \frac{1}{cosθ} - \frac{1}{(secθ + tanθ)}$

$LHS= \frac{1}{secθ - tanθ} - \frac{1}{cosθ} = \frac{(secθ + tanθ)}{(secθ - tanθ) (secθ + tanθ)} - secθ (\begin{array}{l} Multipying the numerator and \\ denominator by (secθ + tanθ) \end{array}) = \frac{secθ + tanθ}{\sec^{2} θ - \tan^{2} θ} - secθ =sec θ +tan θ - secθ [∵ \sec^{2} θ - \tan^{2} θ = 1] = tanθ RHS= \frac{1}{cosθ} - \frac{1}{secθ + tanθ} =sec θ - \frac{(secθ - tanθ)}{\sec^{2} θ - \tan^{2} θ} (\begin{array}{l} Multipying the numerator and \\ denomenator by (secθ - tanθ) \end{array}) = secθ + tanθ - secθ [∵ \sec^{2} θ - \tan^{2} θ = 1] = tanθ ∴ LHS=RHS Hence Proved$

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Question 18:

Prove that $\frac{(\sin A - 2 \sin^{3} A)}{(2 \cos^{3} A - \cos A)} = \tan A .$

Answer:

$LHS= \frac{(\sin A - 2 \sin^{3} A)}{(2 \cos^{3} A - \cos A)} = \frac{\sin A (1 - 2 \sin^{2} A)}{\cos A (2 \cos^{2} A - 1)} = \tan A {\frac{(\sin^{2} A + \cos^{2} A - 2 \sin^{2} A)}{2 \cos^{2} A - \sin^{2} A - \cos^{2} A}} [∵ \sin^{2} A + \cos^{2} A = 1] = \tan A {\frac{(\cos^{2} A - \sin^{2} A)}{(\cos^{2} A - \sin^{2} A)}} = \tan A =RHS$

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Question 19:

Prove that $\frac{\tan A}{(1 - \cot A)} + \frac{\cot A}{(1 - \tan A)} = (1 + \tan A + \cot A) .$

Answer:

$LHS= \frac{\tan A}{(1 - \cot A)} + \frac{\cot A}{(1 - \tan A)} = \frac{\tan A}{(1 - \cot A)} + \frac{\cot^{2} A}{(\cot A - 1)} [∵ \tan A = \frac{1}{\cot A}] = \frac{\tan A}{(1 - \cot A)} - \frac{\cot^{2} A}{(1 - \cot A)} = \frac{\tan A - \cot^{2} A}{(1 - \cot A)} = \frac{(\frac{1}{\cot A}) - \cot^{2} A}{(1 - \cot A)} = \frac{1 - \cot^{3} A}{\cot A (1 - \cot A)} = \frac{(1 - \cot A) (1 + \cot A + \cot^{2} A)}{\cot A (1 - \cot A)} [∵ a^{3} - b^{3} = (a - b) (a^{2} + a b + b^{2})] = \frac{1}{\cot A} + \frac{\cot^{2} A}{\cot A} + \frac{\cot A}{\cot A} =1+ \tan A + \cot A =RHS Hence proved$

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Question 20:

If sec 5 A = cosec (A − 36) and 5 A is an acute angle, find the value of A.

Answer:

$Given: sec5 A = \cos e c (A - 36^{0}) => cosec (90^{0} - 5 A) = \cos e c (A - 36^{0}) [∵ \cos e c (90^{0} - θ) = \sec θ] => 90^{0} - 5 A = A - 36^{0} =>6 A = 90^{0} + 36^{0} => 6 A = 126^{0} => A = 21^{0}$

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