Rs Aggarwal 2019 Solutions for Class 10 Math Chapter 11 T Ratios Of Some Particular Angles are provided here with simple step-by-step explanations. These solutions for T Ratios Of Some Particular Angles are extremely popular among Class 10 students for Math T Ratios Of Some Particular Angles 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 11 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.

#### Question 1:

sin 60° cos 30° + cos 60° sin 30°

On substituting the values of various T-ratios, we get:
sin 60o cos 30o + cos 60o sin 30o
$=\left(\frac{\sqrt{3}}{2}×\frac{\sqrt{3}}{2}+\frac{1}{2}×\frac{1}{2}\right)=\left(\frac{3}{4}+\frac{1}{4}\right)=\frac{4}{4}=1$

#### Question 2:

cos 60° cos 30° − sin 60° sin 30°

On substituting the values of various T-ratios, we get:
cos 60o cos 30o − sin 60o sin 30o
$=\left(\frac{1}{2}×\frac{\sqrt{3}}{2}-\frac{\sqrt{3}}{2}×\frac{1}{2}\right)=\left(\frac{\sqrt{3}}{4}-\frac{\sqrt{3}}{4}\right)=0$

#### Question 3:

cos 45° cos 30° + sin 45° sin 30°

On substituting the values of various T-ratios, we get:
cos 45o cos 30o + sin 45o  sin 30o

#### Question 4:

Evaluate:

$\frac{\mathrm{sin}30°}{\mathrm{cos}45°}+\frac{\mathrm{cot}45°}{\mathrm{sec}60°}-\frac{\mathrm{sin}60°}{\mathrm{tan}45°}+\frac{\mathrm{cos}30°}{\mathrm{sin}90°}$

$\frac{\mathrm{sin}30°}{\mathrm{cos}45°}+\frac{\mathrm{cot}45°}{\mathrm{sec}60°}-\frac{\mathrm{sin}60°}{\mathrm{tan}45°}+\frac{\mathrm{cos}30°}{\mathrm{sin}90°}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{1}{2}\right)}{\left(\frac{1}{\sqrt{2}}\right)}+\frac{1}{2}-\frac{\left(\frac{\sqrt{3}}{2}\right)}{1}+\frac{\left(\frac{\sqrt{3}}{2}\right)}{1}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{2}}{2}+\frac{1}{2}-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\phantom{\rule{0ex}{0ex}}=\frac{\sqrt{2}+1}{2}$

#### Question 5:

Evaluate:

$\frac{5{\mathrm{cos}}^{2}60°+4{\mathrm{sec}}^{2}30°-{\mathrm{tan}}^{2}45°}{{\mathrm{sin}}^{2}30°+{\mathrm{cos}}^{2}30°}$

$\frac{5{\mathrm{cos}}^{2}60°+4{\mathrm{sec}}^{2}30°-{\mathrm{tan}}^{2}45°}{{\mathrm{sin}}^{2}30°+{\mathrm{cos}}^{2}30°}\phantom{\rule{0ex}{0ex}}=\frac{5{\left(\frac{1}{2}\right)}^{2}+4{\left(\frac{2}{\sqrt{3}}\right)}^{2}-{\left(1\right)}^{2}}{{\left(\frac{1}{2}\right)}^{2}+{\left(\frac{\sqrt{3}}{2}\right)}^{2}}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{5}{4}+\frac{4×4}{3}-1\right)}{\left(\frac{1}{4}+\frac{3}{4}\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{5}{4}+\frac{16}{3}-\frac{1}{1}\right)}{\left(\frac{4}{4}\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{15+64-12}{12}\right)}{\left(\frac{4}{4}\right)}\phantom{\rule{0ex}{0ex}}=\frac{\left(\frac{67}{12}\right)}{\left(1\right)}\phantom{\rule{0ex}{0ex}}=\frac{67}{12}$

#### Question 6:

2 cos2 60° + 3 sin2 45° − 3 sin2 30° + 2 cos2 90°

On substituting the values of various T-ratios, we get:
2 cos2 60o + 3 sin2 45o − 3 sin2 30o + 2 cos2 90o

#### Question 7:

cot230° − 2cos230° − $\frac{3}{4}$ sec245° + $\frac{1}{4}$ cosec230°

On substituting the values of various T-ratios, we get:
cot2 30o − 2 cos2 30o − $\frac{3}{4}$ sec2 45o + $\frac{1}{4}$ cosec2 30o

#### Question 8:

(sin230° + 4cot245° − sec260°)(cosec245° sec230°)

On substituting the values of various T-ratios, we get:
(sin2 30o + 4 cot2 45o − sec2 60o )(cosec2 45o sec2 30o)

#### Question 9:

$\frac{4}{{\mathrm{cot}}^{2}30°}+\frac{1}{{\mathrm{sin}}^{2}30°}-2{\mathrm{cos}}^{2}45°-{\mathrm{sin}}^{2}0°$

On substituting the values of various T-ratios, we get:

#### Question 10:

Show that:
(i) $\frac{1-\mathrm{sin}60°}{\mathrm{cos}60°}=\frac{\mathrm{tan}60°-1}{\mathrm{tan}60°+1}$
(ii) $\frac{\mathrm{cos}30°+\mathrm{sin}60°}{1+\mathrm{sin}30°+\mathrm{cos}60°}=\mathrm{cos}30°$

(i)

Hence, LHS = RHS

(ii)

Hence, LHS = RHS

1sin60°cos60°

#### Question 11:

Verify each of the following:
(i) sin 60° cos 30° − cos 60° sin 30° = sin 30°
(ii) cos 60° cos 30° + sin 60° sin 30° = cos 30°
(iii) 2 sin 30° cos 30° = sin 60°
(iv) 2 sin 45° cos 45° = sin 90°

(i) sin 60o cos 30o − cos 60o sin 30o

∴ sin 60o cos 30o − cos 60o sin 30o = sin 30o

(ii) cos 60o cos 30o + sin 60o sin 30o

∴ cos 60o cos 30o + sin 60o sin 30o = cos 30o

(iii) 2 sin 30o cos 30o

∴ 2 sin 30o cos 30o = sin 60o

(iv) 2 sin 45o cos 45o
$=2×\frac{1}{\sqrt{2}}×\frac{1}{\sqrt{2}}=1$
Also, sin 90o = 1
∴ 2 sin 45o cos 45o = sin 90o

#### Question 12:

If A = 45°, verify that:
(i) sin 2A = 2 sin A cos A
(ii) cos 2A = 2 cos2 A − 1 = 1 − 2 sin2 A

A = 45o
⇒ 2A = 2 $×$ 45o = 90o

(i) sin 2A = sin 90o = 1
2 sin A cos A = 2 sin 45o cos 45o
∴ sin 2A = 2 sin A cos A

(ii) cos 2A = cos 90o = 0
2 cos2 A − 1 = 2 cos2 45o − 1 =
Now, 1 − 2 sin2 A
∴ cos 2A 2 cos2 A − 1 = 1 − 2 sin2 A

#### Question 13:

If A = 30°, verify that:
(i) $\mathrm{sin}2A=\frac{2\mathrm{tan}A}{1+{\mathrm{tan}}^{2}A}$
(ii) $\mathrm{cos}2A=\frac{1-{\mathrm{tan}}^{2}A}{1+{\mathrm{tan}}^{2}A}$
(iii) $\mathrm{tan}2A=\frac{2\mathrm{tan}A}{1-{\mathrm{tan}}^{2}A}$

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

​(i) sin 2A = sin 60o = $\frac{\sqrt{3}}{2}$

(ii) cos 2A = cos 60o = $\frac{1}{2}$

(iii) tan 2A = tan 60o = $\sqrt{3}$

=2tanA1+tan2A

#### Question 14:

If A = 60° and B = 30°, verify that:
(i) sin (A + B) = sin A cos B + cos A sin B
(ii) cos (A + B) = cos A cos B − sin A sin B

A = 60o and B = 30o
Now, A + B = 60o + 30o​ = 90o
Also, A − B = 60o − 30o = 30o

(i) sin (A + B) = sin 90o = 1
sin A cos B + cos A sin B = sin 60o cos 30o + cos 60o sin 30o

∴ sin (A + B) = sin A cos B + cos A sin B

(ii) cos (A + B) = cos 90o = 0
cos A cos B − sin A sin B =cos 60o cos 30o − sin 60o sin 30o
∴​ cos (A + B) = cos A cos B − sin A sin B

#### Question 15:

If A = 60° and B = 30°, verify that:

(i) sin (A B) = sin A cos B − cos A sin B
(ii) cos (AB) = cos A cos B + sin A sin B
(iii) tan (A − B) =

(i) sin (A − B) = sin 30o = $\frac{1}{2}$
sin A cos B − cos A sin B = sin 60o cos 30o − cos 60o sin 30o

∴ sin (A − B) = sin A cos B − cos A sin B

(ii) cos (AB) = cos 30o = $\frac{\sqrt{3}}{2}$
cos A cos B + sin A sin B = cos 60o cos 30o + sin 60o sin 30o
=
∴​ cos (AB) = cos A cos B + sin A sin B

(iii) tan (AB) = tan 30o = $\frac{1}{\sqrt{3}}$

∴​ tan (AB) =

#### Question 16:

If A and B are acute angles such that tan A = $\frac{1}{3}$, tan B$\frac{1}{2}$ and tan (A + B) =  , show that A + B = 45°.

Given:

13

#### Question 17:

Using the formula, tan , find the value of tan 60°, it being given that tan 30° = $\frac{1}{\sqrt{3}}$.

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

By substituting the value of the given T-ratio, we get:

∴ tan 60o = $\sqrt{3}$

#### Question 18:

Using the formula,  , find the value of cos 30°, it being given that cos 60° = $\frac{1}{2}$.

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

By substituting the value of the given T-ratio, we get:

∴ cos 30o = $\frac{\sqrt{3}}{2}$

#### Question 19:

Using the formula,  , find the value of sin 30°, it being given that cos 60° = $\frac{1}{2}$.

A = 30o
⇒ 2A = 2 $×$ 30o = 60o

By substituting the value of the given T-ratio, we get:

∴ sin 30o = $\frac{1}{2}$

#### Question 20:

In the adjoining figure, ∆ABC is a right-angled triangle in which ∠B = 90°, ∠A = 30° and AC = 20 cm.
Find (i) BC, (ii) AB.

From the given right-angled triangle, we have:

#### Question 21:

In the adjoining figure, ∆ABC is right-angled at B and ∠A = 30°. If BC = 6 cm, find (i) AB, (ii) AC.

From the given right-angled triangle, we have:

#### Question 22:

In the adjoining figure, ∆ABC is a right-angled at B and ∠A = 45°. If AC$3\sqrt{2}$ cm,
find (i) BC, (ii) AB.

From right-angled ∆ABC, we have:

#### Question 23:

If sin (A + B) = 1 and cos (A − B) = 1, 0° ≤ (A + B) ≤ 90° and A > B, then find A and B.

Here, sin (A + B) = 1
⇒ sin (A+ B) = sin 90o                     [∵ sin 90o = 1]
A + B = 90o​                       ...(i)

Also, cos (AB) = 1
⇒​ cos (A − B) = cos 0o                    [∵​ cos 0o = 1]
A B = 0o                        ...(ii)

Solving (i) and (ii), we get:
A = 45o and B = 45o

#### Question 24:

If sin (AB) = $\frac{1}{2}$ and cos (A + B) = $\frac{1}{2}$, 0° < (A + B) < 90° and A > B, then find A and B.

Here, sin (A − B) = $\frac{1}{2}$
⇒ sin (A B) = sin 30o                [∵ sin 30o = $\frac{1}{2}$]
A − B = 30o​                                  ...(i)

Also, cos (A + B) = $\frac{1}{2}$
⇒​ cos (A + B) =  cos 60o              [∵​ cos 60o = $\frac{1}{2}$]
A + B = 60o                                 ...(ii)

Solving (i) and (ii), we get:
A = 45o and B = 15o

#### Question 25:

If tan (A − B) = $\frac{1}{\sqrt{3}}$ and tan (A + B) = $\sqrt{3}$, 0° < (A + B) < 90° and A > B, then find A and B.

Here, tan (A − B) = $\frac{1}{\sqrt{3}}$
⇒ tan (A B) = tan 30o       [∵ tan 30o = $\frac{1}{\sqrt{3}}$]
A − B = 30o                     ...(i)

Also, tan (A + B) = $\sqrt{3}$
⇒​ tan (A + B) =  tan 60o        [∵ tan 60o = $\sqrt{3}$]
A + B = 60o                           ...(ii)

Solving (i) and (ii), we get:
A = 45o and B = 15o

#### Question 26:

If $3x=\mathrm{cosec}\theta$ and $\frac{3}{x}=\mathrm{cot}\theta$, find the value of $3\left({x}^{2}-\frac{1}{{x}^{2}}\right)$.             [CBSE 2010]

$3\left({x}^{2}-\frac{1}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{9}{3}\left({x}^{2}-\frac{1}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left(9{x}^{2}-\frac{9}{{x}^{2}}\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left[{\left(3x\right)}^{2}-{\left(\frac{3}{x}\right)}^{2}\right]$
$=\frac{1}{3}\left[{\left(\mathrm{cosec}\theta \right)}^{2}-{\left(\mathrm{cot}\theta \right)}^{2}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left({\mathrm{cosec}}^{2}\theta -{\mathrm{cot}}^{2}\theta \right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}\left(1\right)\phantom{\rule{0ex}{0ex}}=\frac{1}{3}$

#### Question 27:

If and , find the values of .

Disclaimer: $\mathrm{cos}15°$ can also be calculated by taking .