NCERT Solutions for Class 10 Maths Chapter 8 - Introduction To Trigonometry And Its Applications

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Answer:

Given: cos A = $\frac{4}{5}$                                                  .....(1)
$\tan \mathrm{A}=\frac{\sin \mathrm{A}}{\cos \mathrm{A}}$
We have the value of cos A, we need to find the value of sin A.
Also we know that, $\sin \mathrm{A}=\sqrt{1-{\cos }^2\mathrm{A}}$           .....(2)               (∵ sin² θ +cos² θ = 1)

Substituting (1) in (2), we get
$\begin{array}{rcl}\mathrm{sinA}& =& \sqrt{1-\frac{16}{25}}\\ & =& \sqrt{\frac{9}{25}}\\ & =& \frac{3}{5}\end{array}$

Therefore, 
tan A = $\frac{{\displaystyle \frac{3}{5}}}{{\displaystyle \frac{4}{5}}}=\frac{3}{4}$

Hence, the correct answer is Option(B).

Answer:

Given: $\sin \mathrm{A}=\frac{1}{2}$                                                                      .....(1)
And we know that, $\cot \mathrm{A}=\frac{1}{\tan \mathrm{A}}=\frac{\cos \mathrm{A}}{\sin \mathrm{A}}$                             .....(2)
We need to find the value of cos A.
$\cos \mathrm{A}=\sqrt{1-{\sin }^2\mathrm{A}}$                                                                 .....(3)                  (∵ sin²θ +cos²θ = 1)

Substituting (1) in (3), we get
cos A = $\sqrt{1-\frac{1}{4}}$ = $\sqrt{\frac{3}{4}}$ =$\frac{\sqrt{3}}{2}$

Substituting values of sin A and cos A in (2), we get
$\cot \mathrm{A}=\frac{{\displaystyle \frac{\sqrt{3}}{2}}}{{\displaystyle \frac{1}{2}}}=\sqrt{3}$
Hence, the correct answer is Option(A).

Answer:

We have,cosec (75° + θ) – sec(15°$-$ θ) $-$ tan(55° + θ) + cot(35°$-$ θ)

= cosec[90°$-$(15°$-$ θ)] – sec(15°$-$ θ) – tan(55° + θ) + cot[ 90°$-$ (55° + θ)]

= sec(15°$-$ θ) – sec(15°$-$ θ) $-$ tan(55° + θ) + tan(55° + θ)             (∵cosec (90°$-$ θ) = sec θ and cot(90°$-$ θ) = tan θ)
= 0

Hence, the correct answer is Option(B).

Answer:

Given : $\sin \theta =\frac{a}{b}$
We know that, sin²θ + cos²θ =1
⇒ cos²θ = 1 $-$ sin²θ
$\begin{array}{rcl}& \Rightarrow & \cos \theta =\sqrt{1-{\sin }^2\theta }\\ & =& \sqrt{1-{\left(\frac{a}{b}\right)}^2}\\ & =& \frac{\sqrt{b^2-a^2}}{b}\end{array}$

Hence, the correct answer is Option(C).

Answer:

Given: cos(α + β) = 0
We can write, cos(α + β) = cos 90° (∵  cos 90° = 0)
By comparing cosine equation on either sides,
We get(α + β) = 90°
⇒ α = 90° $-$ β
Now we need to reduce sin (α $-$ β )
So, sin(α $-$ β)
= sin(90°$-$ β $-$ β)                  (∵ α = 90°$-$ β)
= sin(90°$-$2β)
= cos 2β                                  (∵ sin(90°$-$ θ) = cos θ)

Therefore, sin(α $-$ β) = cos 2β
Hence, the correct answer is Option(B).

Answer:

The value of tan 1°. tan 2°.tan 3° …… tan 89°

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.tan 45°.tan 46°.tan 47°…tan 87°.tan 88°.tan 89°

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.tan 46°.tan 47°…tan 87°.tan 88°.tan 89°                     (∵ tan 45° = 1)

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.tan(90° $-$ 44°).tan(90° $-$ 43°)…tan(90° $-$ 3°). tan(90°$-$ 2°).tan(90°$-$1°)

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.cot 44°.cot 43°…cot 3°.cot 2°.cot 1°                             (∵ tan(90°$-$ θ) = cot θ)
= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.$\frac{1}{\tan 44°}.\frac{1}{\tan 43°}....\frac{1}{\tan 3°}.\frac{1}{\tan 2°}.\frac{1}{\tan 1°}$
= 1
Hence, the correct answer is Option (B).

Answer:

Given: cos 9$\alpha$ = sin $\alpha$ and 9$\alpha$ < 90° i.e. 9$\alpha$ is an acute angle

And we know that, sin(90°$-$ θ) = cos θ by property.
So, we can write cosine in terms of sine using this property,
cos 9$\alpha$ = sin (90° $-$ 9$\alpha$)
Thus, sin (90° $-$ 9$\alpha$) = sin$\alpha$                                (∵cos 9$\alpha$ = sin(90° $-$ 9$\alpha$) and sin(90°$-$ $\alpha$) = sin$\alpha$ )
⇒ 90°$-$9$\alpha$ = $\alpha$
⇒ 10$\alpha$ = 90°              (By rearranging)
⇒ $\alpha$ = 9°
We have got the value of $\alpha$ i.e. $\alpha$ = 9°
Putting it in tan 5$\alpha$, we get
tan 5$\alpha$ = tan ($5\times 9$) = tan 45° = 1
∴ tan 5$\alpha$ = 1

Hence, the correct answer is Option(C).

Answer:

Given: ∠C = 90°
By the property of triangle, the sum of the three angles is equal to 180$°$
∠A + ∠B + ∠C = 180°
⇒ ∠A + ∠B + 90° = 180°           (∵ ∠C = 90° )
⇒ ∠A + ∠B = 90°

Thus, cos(A + B) = cos 90° = 0

Hence, the correct answer is Option(A).

Answer:

Given: sin A + sin² A = 1
⇒ sin A = 1 – sin² A
⇒ sin A = cos² A                                  (∵ sin² θ +cos² θ = 1 )
Squaring both sides, we get
⇒ sin² A = cos⁴ A or
⇒cos⁴ A = sin² A
⇒cos⁴ A = 1 – cos² A
∴ cos² A + cos⁴ A = 1
Hence, the correct answer is Option(A).

Answer:

Given: $\sin \alpha =\frac{1}{2} \mathrm{and} \cos \beta =\frac{1}{2}$

sin α = $\frac{1}{2}$ = sin 30° and cos β = $\frac{1}{2}$ = cos 60°
Comparing each sine and cosine angles respectively, we get
α = 30° and β = 60°

Thus, α + β = 30° + 60° =  90°

Hence, the correct answer is Option(D).    

Answer:

Given, $\frac{{\sin }^{2}22°+{\sin }^{2}68°}{{\cos }^{2}22°+{\cos }^{2}68°}+{\sin }^{2}63°+\cos 63°\sin 27°$

= $\frac{{\sin }^2 22°+{\sin }^2 68°}{{\cos }^2 22°+{\cos }^2 68°}+{\sin }^2 63°+\cos 63° \sin \left(90-63\right)°$                         

= $\frac{{\sin }^2 22°+{\sin }^2 68°}{{\cos }^2 22°+{\cos }^2 68°}+{\sin }^2 63°+\cos 63°\cos 63°$                           [$\sin (90-\theta ) = \cos \theta$ ]

= $\frac{{\sin }^2 22°+{\sin }^2 68°}{{\cos }^2 22°+{\cos }^2 68°}+{\sin }^2 63°+{\cos }^2 63°$

=  $\frac{{\sin }^2 22°+{\sin }^2 68°}{{\cos }^2 22°+{\cos }^2 68°}+1$                                                            [ ${\sin }^2\theta + {\cos }^2\theta = 1$ ]

=  $\frac{{\sin }^2 22°+{\cos }^2 22°}{{\cos }^2 22°+{\sin }^2 22°}+1$                                                  [$\sin (90-\theta ) = \cos \theta$ ]

= $\frac{1}{1}+1$ 
= 2 

Hence, the correct answer is Option(B).

Answer:

Given: 4 tan θ = 3                                                           .....(1)

We have to evaluate, $\left(\frac{4 \sin \theta -\cos \theta }{4\sin \theta +\cos \theta }\right)$

Dividing numerator and denominator by cos θ and substituting (1), we get 

$\begin{array}{rcl} \frac{{\displaystyle \frac{4 \sin \theta -\cos \theta }{\cos \theta }}}{{\displaystyle \frac{4\sin \theta +\cos \theta }{\cos \theta }}}& =& \frac{{\displaystyle \frac{4 \sin \theta }{\cos \theta }-\frac{\cos \theta }{\cos \theta }}}{{\displaystyle \frac{{\displaystyle 4 \sin \theta }}{{\displaystyle \cos \theta }}+\frac{\cos \theta }{\cos \theta }}}\\ & =& \frac{4\tan \theta -1}{4\tan \theta +1}\\ & =& \frac{4\times {\displaystyle \frac{3}{4}}-1}{4\times \frac{3}{4}+1}\\ & =& \frac{2}{4}\\ & =& \frac{1}{2}\end{array}$

Hence, the correct answer is Option(C).

Answer:

Given: sinθ – cosθ = 0
$\Rightarrow \sin \theta = \cos \theta$
$\Rightarrow \frac{\sin \theta }{\cos \theta } = 1$
$\Rightarrow \tan \theta = 1$
And we know, tan 45° = 1
So, tan θ = 1 = tan 45°
By comparing above equation, we get θ = 45°
$\begin{array}{rcl}\therefore {\sin }^4\theta + {\cos }^4\theta & =& {\sin }^4\left(45\right)°+{\cos }^4\left(45\right)°\\ & =& {\left(\frac{1}{\sqrt{2}}\right)}^4+{\left(\frac{1}{\sqrt{2}}\right)}^4\\ & =& \frac{1}{4}+\frac{1}{4}\\ & =& \frac{1}{2}\end{array}$

Hence, the correct answer is Option(C).

Answer:

As we know that, sin(90°$-$θ) = cos θ by sine property.

So, sin(45° + θ) $-$ cos(45°$-$ θ)

= sin[90°$-$ ( 45°$-$ θ)] $-$ cos(45°$-$ θ)

= cos(45°$-$ θ) $-$ cos(45°$-$ θ)                                (By using identity)

= 0

Hence, the correct answer is Option (B).

Answer:

Let's assume AB = 2√3 m, i.e. length of shadow on the ground from the pole and BC = 6 m, i.e. height of the pole as mentioned in the diagram.


Let sun make an angle of θ on the ground, such that ∠CAB = θ


So in ∆ABC, tan θ = $\frac{\mathrm{BC}}{\mathrm{AB}}$       (∵ tan θ = $\frac{\mathrm{perpendicular}}{\mathrm{base}}$)

$\Rightarrow \tan \theta = \frac{6}{2\sqrt{3}}$

⇒ tan θ = tan 60° (∵ tan 60° = $\sqrt{3}$)
By comparing, we get
θ = 60°
Thus, the sun’s elevation is 60°.

Hence, the correct answer is Option(A).

Answer:

True

$$\begin{gathered} \frac{\tan 47°}{\cot 43°}=\frac{\tan (90-43)°}{\cot 43°} \\ =\frac{\cot 43°}{\cot 43°} \\ = 1 \end{gathered}$$  (∵ tan (90°$-$ θ) = cot θ)

Answer:

False

cos² 23° – sin² 67° = (cos 23° + sin 67°) (cos 23° – sin 67°)                                  [∵  a² – b²) = (a + b)(a – b)]

= [cos 23° + sin(90°– 23°)] [cos 23°– sin(90°– 23°)]

= (cos 23°+ cos 23°)(cos 23° – cos 23°)                                                                  [∵ sin(90°$-$θ) = cos θ ]

= (cos 23° + cos 23°).0

= 0, which is neither positive nor negative.

Answer:

False

We know,
sin θ increases when 0° ≤ θ ≤ 90°
cos θ decreases when 0° ≤ θ ≤ 90°

And (sin 80°$-$ cos 80°) = (increasing value $-$ decreasing value) which is always equal to a positive value.
∴ (sin 80°$-$ cos 80°) > 0, which is positive (not negative).

Answer:

LHS = $\sqrt{\left(1-{\cos }^2\theta \right) {\sec }^2\theta }$

= $\sqrt{\left({\sin }^2\theta \right){\sec }^2\theta }$                                    $\left({\sin }^2\theta +{\cos }^2\theta = 1\right)$

= $\sqrt{\frac{\left({\sin }^2\theta \right)}{\left({\cos }^2\theta \right)}}$                                          $\left(\because {\sec }^2\theta = \frac{1}{{\cos }^2\theta }\right)$
= $\frac{\sin \theta }{\cos \theta }$
= $\tan \theta$

= RHS

Hence, the given expression is True.

Answer:

True

Given: cos A + cos²A = 1
⇒ cos A = 1$-$ cos²A                                    (∵ sin²θ  + cos²θ = 1 ⇒ sin²θ = 1$-$ cos²θ)
⇒ cos A = sin²A                                           .....(1)
Squaring both sides,
we get cos²A = sin⁴A                                  .....(2)

We have to find whether sin²A + sin⁴ A = 1

So, adding (1) and (2), we get

sin²A + sin⁴ A = cos A + cos²A                                    (As given)
∴ sin²A+ sin⁴ A = 1

Hence, the given statement is True.

Answer:

False

L.H.S= (tan θ + 2) (2 tan θ + 1)

Multiplying them,

= 2 tan²θ + tan θ + 4 tan θ + 2

= 2 tan²θ + 5 tanθ + 2

= 2(sec²θ$-$1) + 5tanθ + 2                                 (∵ sec²θ $-$ tan²θ = 1 ⇒ tan²θ = sec²θ$-$1)
= 2 sec²θ $-$ 2 + 5 tan θ + 2
= 5 tan θ + 2 sec² θ ≠R.H.S
∴, L.H.S ≠ R.H.S

Hence, the given expression is False.

Answer:

False

Let us understand this by an example: A tower of 30 m high casts a shadow $10\sqrt{3}$ m long on the ground, then the sun’s elevation is 60°.

$$\begin{gathered} \mathrm{In} ∆\mathrm{ACB}, \\ \tan \theta =\frac{\mathrm{BC}}{\mathrm{AB}}=\frac{30}{10\sqrt{3}} \left(\tan \theta =\frac{\mathrm{Perpendicular}}{\mathrm{Base}}\right) \\ \Rightarrow \tan \theta =\frac{3}{\sqrt{3}}=\sqrt{3} \end{gathered}$$     
⇒ tan θ = tan 60°
∴ θ = 60°

Now, if the same height of the tower casts a shadow $20\sqrt{3}$ m more than that from the preceding length, then the Sun’s elevation becomes 30°.


$$\begin{gathered} \mathrm{In} ∆\mathrm{APB}, \\ \tan \theta =\frac{\mathrm{AB}}{\mathrm{PB}}=\frac{\mathrm{AB}}{\mathrm{PC}+\mathrm{CB}} \end{gathered}$$ 

$\begin{array}{rcl}& \Rightarrow & \tan \theta =\frac{30}{10\sqrt{3}+20\sqrt{3}}\\ & =& \frac{30}{30\sqrt{3}}\\ & =& \frac{1}{\sqrt{3}}\\ & =& \tan 30°\end{array}$
∴ θ = 30°

Hence, we can conclude that as the length of the shadow of the tower increases, the angle of elevation of the sun decreases. And the above statement is false.

Answer:

False

Let P be the point where the man is standing, C be the point of cloud and the height of the cloud from the surface of the platform is h.
The angles, θ₁ = the angle of elevation of the cloud and θ₂ = the angle of depression of the cloud.

The height of reflection of cloud is h + 3 because height of lake is also added to the platform’s height.

So, the angle of depression is different in lake to the angle of elevation of cloud above the surface of the lake.

In $∆\mathrm{MPC}$,

$\tan {\theta }_1=\frac{\mathrm{CM}}{\mathrm{PM}} = \frac{h}{\mathrm{PM}}$
$\Rightarrow \frac{\tan {\theta }_1}{h}=\frac{1}{\mathrm{PM}}$                                                                    .....(1)

In $∆\mathrm{PMD}$,

$\tan {\theta }_2=\frac{\mathrm{MD}}{\mathrm{PM}} = \frac{\mathrm{MO}+\mathrm{OD}}{\mathrm{PM}}=\frac{3+h}{\mathrm{PM}}$
$\Rightarrow \frac{\tan {\theta }_2}{3+h}=\frac{1}{\mathrm{PM}}$                                                                   .....(2)

From (1) and (2) we get,

$\Rightarrow \frac{\tan {\theta }_1}{h}=\frac{\tan {\theta }_2}{3+h}$
$\Rightarrow {\theta }_1\ne {\theta }_2$

∴ This proves that the angle of elevation is not equal to the angle of depression of the sun.

Answer:

False

Given: a is a positive number and a ≠ 1

⇒ AM ≥ GM
(Arithmetic Mean (AM) of a list of non-negative real numbers is greater than or equal to the Geometric mean (GM) of the same list)

If a and b be such numbers, then

$AM=\frac{a+b}{2}\text{ and GM }= \sqrt{ab}$

By assuming that  $a+\frac{1}{a}$= 2sinθ  is true 
Applying the inequality on $a \mathrm{and} \frac{1}{a}$ , we get 
$$\begin{gathered} \Rightarrow \frac{a+{\displaystyle \frac{1}{a}}}{2}\ge \sqrt{\left(a\times \frac{1}{a}\right)} \\ \Rightarrow \frac{a+{\displaystyle \frac{1}{a}}}{2}\ge 1 \\ \Rightarrow a+\frac{1}{a}\ge 2 \end{gathered}$$
$\Rightarrow 2 \sin \theta \ge 2 ( \mathrm{By} \mathrm{our} \mathrm{assumption})$

⇒ sin θ ≥ 1

But $-$1 ≤ sin θ ≤ 1

∴ Our assumption is wrong and that 2 sin θ cannot be equal to $a+\frac{1}{a}$.

Answer:

False

Given: a ≠ b and ab > 0
(Because Arithmetic Mean (AM) of a list of non-negative real numbers is greater than or equal to the Geometric mean (GM) of the same list)
⇒ AM ≥ GM
If a and b be such numbers, then
$\mathrm{AM}=\frac{a+b}{2}$, GM = $\sqrt{ab}$
By assuming that $\cos \theta =\frac{a^2+b^2}{2ab}$   is true statement.

Similarly, AM and GM of a² and b² will be,

$$\begin{gathered} \mathrm{AM}=\frac{a^2+b^2}{2} \\ \mathrm{GM}=\sqrt{a^2{b}^2} \end{gathered}$$

So $\frac{a^2+b^2}{2}\ge \sqrt{a^2\times b^2}$         (By AM and GM property as mentioned earlier in the answer)
$$\begin{gathered} \Rightarrow \frac{a^2+b^2}{2}\ge ab \\ \Rightarrow \frac{a^2+b^2}{2ab}\ge 1 \end{gathered}$$
$\Rightarrow \cos \theta \ge 1$                                                            ( By our assumption)

But this is not possible since −1 ≤ cos θ ≤ 1
Thus, our assumption is wrong.

Answer:

False

Let the height of the tower be h and BC = x m.

In ∆ABC,

$$\begin{gathered} \tan 30°=\frac{\mathrm{AC}}{\mathrm{BC}} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x} .....\left(1\right) \end{gathered}$$

Now, let the height be 2h (doubled).

$$\begin{gathered} \mathrm{In} ∆\mathrm{PQR}, \\ \tan \theta =\frac{\mathrm{PR}}{\mathrm{QR}}=\frac{2h}{x} \end{gathered}$$
$\Rightarrow \tan \theta =\frac{2}{\sqrt{3}}=1.15$
⇒ θ = tan^-1(1.15) < 60°                                     (∵ tan^-1 (1.15) = 49°)

Hence, the angle of elevation is not doubled when height is doubled.

Answer:

True
Let height of the tower be h and distance of the point from its foot is x.
Let angle of elevation be θ₁.

In $∆\mathrm{ABC}$,

$\tan {\theta }_1=\frac{\mathrm{AC}}{\mathrm{BC}}=\frac{h}{x}$
$\Rightarrow {\theta }_1={\tan }^{-1}\left(\frac{h}{x}\right)$                                                               .....(1)
Now, when both(height and distance) are increased by ten percent, we get 
New height = h + 10% of  h _{= $\frac{11}{10}h$}
New distance = x + 10% of  x = $\frac{11}{10}x$

In _{$∆\mathrm{PQR}$,}

$\tan {\theta }_2=\frac{\mathrm{PR}}{\mathrm{QR}}=\frac{{\displaystyle \frac{11h}{10}}}{{\displaystyle \frac{11x}{10}}}$                                                 .....(2)
$\Rightarrow {\theta }_2={\tan }^{-1}\left(\frac{h}{x}\right)$
From (1) and (2)  we get θ₁ = θ₂

Hence, it is true.

Answer:

$\begin{array}{rcl}\mathrm{LHS}& =& \frac{\sin \theta }{1+\cos \theta }+\frac{1+\cos \theta }{\sin \theta }\\ & =& \frac{{\sin }^2\theta +{\left(1+\cos \theta \right)}^2}{\sin \theta \left(1+\cos \theta \right)}\\ & =& \frac{{\sin }^2\theta +1+{\cos }^2\theta +2\cos \theta }{\sin \theta \left(1+\cos \theta \right)} \left[\because {\left(a+b\right)}^2=a^2+b^2+2ab\right]\\ & =& \frac{1+1+2\cos \theta }{\sin \theta \left(1+\cos \theta \right)} \left[\because {\sin }^2\theta +{\cos }^2\theta =1\right]\\ & =& \frac{2\left(1+\cos \theta \right)}{\sin \theta \left(1+\cos \theta \right)}\\ & =& \frac{2}{\sin \theta }\\ & =& 2\cos ec\theta \left[\cos ec\theta =\frac{1}{\sin \theta }\right]\\ & =& \mathrm{RHS}\end{array}$

Hence proved.

Answer:

$\begin{array}{rcl}\mathrm{LHS}& =& \frac{\tan A}{1+\sec A}-\frac{\tan A}{1-\sec A}\\ & =& \frac{\tan A\left(1-\sec A\right)-\tan A\left(1+\sec A\right)}{\left(1+\sec A\right)\left(1-\sec A\right)}\\ & =& \frac{\tan A\left(1-\sec A-1-\sec A\right)}{1-{\sec }^{2}A}\\ & =& \frac{\tan A\left(-2\sec A\right)}{1-{\sec }^{2}A} \left[\because \left(a+b\right)\left(a-b\right)=a^2-b^2\right]\\ & =& \frac{2\tan A \sec A}{{\sec }^{2}A-1} \\ & =& \frac{2\tan A\sec A}{{\tan }^{2}A} \left[\because se{c}^{2}A-{\tan }^{2}A=1\right]\\ & =& \frac{2 \sec A}{\tan A}\\ & =& \frac{{\displaystyle \frac{2}{\cos A}}}{{\displaystyle \frac{\sin A}{\cos A}}}\\ & =& \frac{2}{\sin A}\\ & =& 2\mathrm{cosec}A\\ & =& \mathrm{RHS} \end{array}$

Hence proved.

Answer:

$\tan A=\frac{3}{4}=\frac{\mathrm{Perpendicular}\left(P\right)}{\mathrm{Base}\left(B\right)}$
Let $P=3x \mathrm{and} B=4x$.
By pythagoras theorem
H² = P² + B² = (3x)² + (4x)²
H² = 25x²
∴ H = 5x



$$\begin{gathered} \sin A=\frac{P}{H}=\frac{3x}{5x}=\frac{3}{5} \\ \cos A=\frac{B}{H}=\frac{4x}{5x}=\frac{4}{5} \\ \therefore \sin A \cos A=\frac{3}{5}\times \frac{4}{5}=\frac{12}{25} \end{gathered}$$

Hence proved.

Answer:

$$\begin{gathered} \left(\sin \alpha +\cos \alpha \right)\left(\tan \alpha +\cot \alpha \right) \\ =\left(\sin \alpha +\cos \alpha \right)\left(\frac{\sin \alpha }{\cos \alpha }+\frac{\cos \alpha }{\sin \alpha }\right) \left[\tan \alpha =\frac{\sin \alpha }{\cos \alpha } \mathrm{and} cot\alpha =\frac{\cos \alpha }{\sin \alpha }\right] \\ =\left(\sin \alpha +\cos \alpha \right)\left(\frac{{\sin }^2\alpha +{\cos }^2\alpha }{\sin \alpha \cos \alpha }\right) \\ =\left(\sin \alpha +\cos \alpha \right)\left(\frac{1}{\sin \alpha \cos \alpha }\right) \left[\because {\sin }^2\alpha +{\cos }^2\alpha =1\right] \\ =\frac{1}{\cos \alpha }+\frac{1}{\sin \alpha } \left[sec\alpha =\frac{1}{\cos \alpha } \mathrm{and} \cos ec\alpha =\frac{1}{\sin \alpha }\right] \\ =\sec \alpha +\mathrm{cosec}\alpha \\ =\mathrm{RHS} \end{gathered}$$

Hence proved.

Answer:

$\begin{array}{rcl}\mathrm{RHS}& =& {\tan }^3{60}^{\mathrm{o}}-2\sin {60}^{\mathrm{o}}\\ & =& {\left(\sqrt{3}\right)}^3-2\times \frac{\sqrt{3}}{2}\\ & =& 3\sqrt{3}-\sqrt{3}\\ & =& 2\sqrt{3}\\ & & \\ \mathrm{LHS}& =& \left(\sqrt{3}+1\right)\left(3-\cot {30}^{\mathrm{o}}\right)\\ & =& \left(\sqrt{3}+1\right)\times \left(3-\sqrt{3}\right)\\ & =& \left(\sqrt{3}+1\right)\times \sqrt{3}\times \left(\sqrt{3}-1\right)\\ & =& \sqrt{3}\left({\sqrt{3}}^2-1\right)\\ & =& 2\sqrt{3}\end{array}$

Thus, LHS = RHS.

Hence proved.

Answer:

$\begin{array}{rcl}\mathrm{LHS}& =& 1+\frac{{\cot }^2\alpha }{1+\mathrm{cosec}\alpha }\\ & =& 1+\frac{{\displaystyle \frac{{\cos }^2\alpha }{{\sin }^2\alpha }}}{1+{\displaystyle \frac{1}{\sin \alpha }}} \left[\because \cot \alpha =\frac{\cos \alpha }{\sin \alpha } \mathrm{and} \mathrm{cosec}\alpha =\frac{1}{\sin \alpha }\right]\\ & =& 1+\frac{{\displaystyle {\cos }^2\alpha }}{\sin \alpha \left(1+\sin \alpha \right)}\\ & =& \frac{{\displaystyle \sin \alpha \left(1+\sin \alpha \right)+{\cos }^2\alpha }}{\sin \alpha \left(1+\sin \alpha \right)}\\ & =& \frac{{\displaystyle \sin \alpha +{\sin }^2\alpha +{\cos }^2\alpha }}{\sin \alpha \left(1+\sin \alpha \right)}\\ & =& \frac{1+\sin \alpha }{\sin \alpha \left(1+\sin \alpha \right)} \left[{\sin }^2\alpha +{\cos }^2\alpha =1\right]\\ & =& \frac{1}{\sin \alpha }\\ & =& \cos ec\alpha \left[\frac{1}{\sin \alpha }=\cos ec\alpha \right]\\ & =& \mathrm{RHS}\end{array}$

Hence proved.

Answer:

$\begin{array}{rcl}\mathrm{LHS}& =& \tan \theta +\tan \left(90°-\theta \right)\\ & =& \tan \theta +\cot \theta \left[\because \tan \left(90°-\theta \right)=\cot \theta \right]\\ & =& \frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta } \left[\because \tan \theta =\frac{\sin \theta }{\cos \theta } \mathrm{and} \cot \theta =\frac{\cos \theta }{\sin \theta } \right]\\ & =& \frac{{\sin }^2\theta +{\cos }^2\theta }{\sin \theta \cos \theta } \left[\because {\sin }^2\theta +{\cos }^2\theta =1\right]\\ & =& \frac{1}{\sin \theta \cos \theta }\\ & =& \sec \theta \mathrm{cosec}\theta \left[\because \sec \theta =\frac{1}{\cos \theta } \mathrm{and} \mathrm{cosec}\theta =\frac{1}{\sin \theta }\right]\\ & =& \sec \theta \sec \left(90°-\theta \right) \left[\because \sec \left(90°-\theta \right)= \mathrm{cosec}\theta \right]\\ & =& \mathrm{RHS}\end{array}$

Hence proved.

Answer:

Let the angle of the sun be $\theta$
Given, the height of the pole is h.



Now in $∆$ ABC,
$$\begin{gathered} \therefore \tan \theta =\frac{\mathrm{AC}}{\mathrm{BC}} \\ \Rightarrow \tan \theta =\frac{h}{\sqrt{3}h} \\ \Rightarrow \tan \theta =\frac{1}{\sqrt{3}} \\ \Rightarrow \tan \theta =\tan 30° \\ \Rightarrow \theta ={30}^o \end{gathered}$$

Thus, the angle of elevation of the sun is 30^o.

Answer:

Given: $\sqrt{3}\tan \theta =1$
$$\begin{gathered} \Rightarrow \tan \theta =\frac{1}{\sqrt{3}} \\ \Rightarrow \tan \theta =\tan 30° \\ \Rightarrow \theta ={30}^{\mathrm{o}} \end{gathered}$$

Now,
$\begin{array}{rcl}{\sin }^2\theta -{\cos }^2\theta & =& {\sin }^2{30}^{\mathrm{o}}-{\cos }^2{30}^{\mathrm{o}}\\ & =& {\left(\frac{1}{2}\right)}^2-{\left(\frac{\sqrt{3}}{2}\right)}^2\\ & =& \frac{1}{4}-\frac{3}{4}\\ & =& \frac{-2}{4}\\ & =& \frac{-1}{2}\end{array}$

Answer:

Given, the height of the ladder = 15m
Let the height of the vertical wall = h 
The ladder makes an angle of 60^o with the wall i.e. θ = 60^o.



$$\begin{gathered} \mathrm{In} ∆\mathrm{QPR}, \\ \cos {60}^{\mathrm{o}}=\frac{\mathrm{PR}}{\mathrm{PQ}}=\frac{h}{15} \\ \Rightarrow \frac{1}{2}=\frac{h}{15} \\ \Rightarrow h=\frac{15}{2} \\ \Rightarrow h=7.5 \mathrm{m} \end{gathered}$$

Thus the required height of the wall is 7.5 meters.

Answer:

$$\begin{gathered} \left(1+{\tan }^2\theta \right)\left(1-\sin \theta \right)\left(1+\sin \theta \right) \\ =\left(1+{\tan }^2\theta \right)\left(1-{\sin }^2\theta \right) \left[\because \left(a-b\right)\left(a+b\right)=a^2-b^2\right] \\ ={\sec }^2\theta \times {\cos }^2\theta \left[\because 1+{\tan }^2\theta =se{c}^2\theta and {\cos }^2\theta +{\sin }^2\theta =1\right] \\ =\frac{1}{{\cos }^2\theta }\times {\cos }^2\theta \left[\because \sec \theta =\frac{1}{\cos \theta }\right] \\ =1 \end{gathered}$$

Answer:

$$\begin{gathered} 2{\sin }^2\theta -{\cos }^2\theta =2 \\ \Rightarrow 2{\sin }^2\theta -\left(1-{\sin }^2\theta \right)=2 \left[\because {\cos }^2\theta +{\sin }^2\theta =1\right] \\ \Rightarrow 2{\sin }^2\theta -1+{\sin }^2\theta =2 \\ \Rightarrow 3{\sin }^2\theta =3 \\ \Rightarrow {\sin }^2\theta =1 \\ \Rightarrow \sin \theta =1 \\ \Rightarrow \sin \theta =\sin {90}^{\mathrm{o}} \\ \Rightarrow \theta ={90}^{\mathrm{o}} \end{gathered}$$

Hence, the value of θ

is 90°.

Answer:

Given, 
$\begin{array}{rcl}\mathrm{LHS}& =& \frac{{\cos }^2\left({45}^{\mathrm{o}}+\theta \right)+{\cos }^2\left({45}^{\mathrm{o}}-\theta \right)}{\tan \left({60}^{\mathrm{o}}+\theta \right).\tan \left({30}^{\mathrm{o}}-\theta \right)}\\ & =& \frac{{\cos }^2\left({45}^{\mathrm{o}}+\theta \right)+\left[\sin \left\{{90}^{\mathrm{o}}-\left({45}^{\mathrm{o}}-\theta \right)\right\}\right]}{\tan \left({60}^{\mathrm{o}}+\theta \right).\cot \left\{{90}^{\mathrm{o}}-\left({30}^{\mathrm{o}}-\theta \right)\right\}} \left[\because \sin \left({90}^{\mathrm{o}}-\theta \right)=\cos \theta \mathrm{and} \cot \left({90}^{\mathrm{o}}-\theta \right)=\tan \theta \right]\\ & =& \frac{{\cos }^2\left({45}^{\mathrm{o}}+\theta \right)+{\sin }^2\left({45}^{\mathrm{o}}+\theta \right)}{\tan \left({60}^{\mathrm{o}}+\theta \right).\cot \left({60}^{\mathrm{o}}+\theta \right)} \left[\because {\cos }^2\left(\theta \right)+{\sin }^2\left(\theta \right)=1\right]\\ & =& \frac{1}{\tan \left({60}^{\mathrm{o}}+\theta \right).{\displaystyle \frac{1}{\tan \left({60}^{\mathrm{o}}+\theta \right)}}} \left[\because cot\left(\theta \right)=\frac{1}{\tan \left(\theta \right)}\right]\\ & =& 1\\ & =& \mathrm{RHS}\end{array}$

Hence proved.

Answer:

Let the angle of elevation of the top of the tower from the eye of the observer be θ.



Given, AB = 22 m , PQ = MB = 1.5 m and QB = PM = 20.5 m
⇒ AM = AB $-$ MB
⇒ AM = 22 $-$ 1.5 = 20.5 m

Now in $∆$APM,
$$\begin{gathered} \tan \theta =\frac{\mathrm{AM}}{\mathrm{PM}}=\frac{20.5}{20.5}=1 \\ \Rightarrow \tan \theta =\tan {45}^{\mathrm{o}} \\ \Rightarrow \theta ={45}^{\mathrm{o}} \end{gathered}$$

Hence, the required angle of elevation of the top of the tower from the eye of the observer is 45^o.

Answer:

$\begin{array}{rcl}\mathrm{LHS}& =& {\tan }^4\theta +{\tan }^2\theta \\ & =& {\tan }^2\theta \left({\tan }^2\theta +1\right)\\ & =& {\tan }^2\theta .{\sec }^2\theta \left[\because {\sec }^2\theta ={\tan }^2\theta +1\right]\\ & =& \left({\sec }^2\theta -1\right).{\sec }^2\theta \left[\because {\tan }^2\theta ={\sec }^2\theta -1\right]\\ & =& {\sec }^4\theta -{\sec }^2\theta \\ & =& \mathrm{RHS}\end{array}$

Hence proved.

Answer:

$$\begin{gathered} \mathrm{cosec}\theta +\cot \theta =p \\ \Rightarrow \frac{1}{\sin \theta }+\frac{\cos \theta }{\sin \theta }=p \left[\mathrm{cosec}\theta =\frac{1}{\sin \theta } \mathrm{and} \cot \theta =\frac{\cos \theta }{\sin \theta }\right] \\ \Rightarrow \frac{1+\cos \theta }{\sin \theta }=p \\ \Rightarrow \frac{{\left(1+\cos \theta \right)}^2}{{\sin }^2\theta }=p^2 \left[\mathrm{Squaring} \mathrm{both} \mathrm{sides}\right] \\ \Rightarrow \frac{1+2\cos \theta +{\cos }^2\theta }{{\sin }^2\theta }=p^2 \end{gathered}$$
Using componendo and dividendo rule,
$$\begin{gathered} \Rightarrow \frac{\left(1+2\cos \theta +{\cos }^2\theta \right)-{\sin }^2\theta }{\left(1+2\cos \theta +{\cos }^2\theta \right)+{\sin }^2\theta }=\frac{p^2-1}{p^2+1} \\ \Rightarrow \frac{\left(1+2\cos \theta +{\cos }^2\theta \right)-\left(1-{\cos }^2\theta \right)}{1+2\cos \theta +\left({\cos }^2\theta +{\sin }^2\theta \right)}=\frac{p^2-1}{p^2+1} \\ \Rightarrow \frac{2\cos \theta +2{\cos }^2\theta }{2+2\cos \theta }=\frac{p^2-1}{p^2+1} \left[\because {\cos }^2\theta +{\sin }^2\theta =1\right] \\ \Rightarrow \frac{\cos \theta +{\cos }^2\theta }{1+\cos \theta }=\frac{p^2-1}{p^2+1} \\ \Rightarrow \frac{\cos \theta \left(1+\cos \theta \right)}{1+\cos \theta }=\frac{p^2-1}{p^2+1} \\ \Rightarrow \cos \theta =\frac{p^2-1}{p^2+1} \end{gathered}$$

Hence proved.

Answer:

$\begin{array}{rcl}\mathrm{LHS}& =& \sqrt{{\sec }^2\theta +{\mathrm{cosec}}^2\theta }\\ & =& \sqrt{\frac{1}{{\cos }^2\theta }+\frac{1}{{\sin }^2\theta }} \left[\because \sec \theta =\frac{1}{\cos \theta } \mathrm{and} \mathrm{cosec}\theta =\frac{1}{\sin \theta }\right]\\ & =& \sqrt{\frac{{\sin }^2\theta +{\cos }^2\theta }{{\cos }^2\theta {\sin }^2\theta }}\\ & =& \sqrt{\frac{1}{{\cos }^2\theta {\sin }^2\theta }} \left[\because {\sin }^2\theta +{\cos }^2\theta =1\right]\\ & =& \frac{1}{\cos \theta \sin \theta }\\ & =& \frac{{\sin }^2\theta }{\cos \theta \sin \theta }+\frac{{\cos }^2\theta }{\cos \theta \sin \theta } \left[\because {\sin }^2\theta +{\cos }^2\theta =1\right]\\ & =& \frac{\sin \theta }{\cos \theta }+\frac{\cos \theta }{\sin \theta } \\ & =& \tan \theta +\cot \theta \\ & =& \mathrm{RHS}\end{array}$

Hence proved.

Answer:

Let the height of the tower be h.
SR = x m, $\angle$PSR = θ, QS = 20 m and $\angle$PQR = 30^o



Now in $∆$PSR,
$$\begin{gathered} \tan \theta =\frac{\mathrm{PR}}{\mathrm{SR}}=\frac{h}{x} \\ \Rightarrow \tan \theta =\frac{h}{x} \\ \Rightarrow x=\frac{h}{\tan \theta } .....\left(1\right) \end{gathered}$$

Now in $∆$PQR,
$$\begin{gathered} \tan {30}^{\mathrm{o}}=\frac{\mathrm{PR}}{\mathrm{QR}}=\frac{\mathrm{PR}}{\mathrm{QS}+\mathrm{SR}} \\ \Rightarrow \tan {30}^{\mathrm{o}}=\frac{h}{20+x} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{20+x} \\ \Rightarrow 20+x=h\sqrt{3} \\ \Rightarrow 20+\frac{h}{\tan \theta }=h\sqrt{3} .....\left(2\right) \left[\mathrm{From} \left(1\right)\right] \end{gathered}$$

After moving 20 m towards the tower the angle of elevation of the top increases by 15^o.
i.e.  $\angle$PSR = θ = $\angle$PQR + 15^o
$$\begin{gathered} \Rightarrow \theta =30°+15°=45° \\ \therefore 20+\frac{h}{\tan 45°}=h\sqrt{3} \left[\mathrm{From} \left(2\right)\right] \\ \Rightarrow 20+\frac{h}{1}=h\sqrt{3} \\ \Rightarrow 20=h\sqrt{3}-h \\ \Rightarrow h\left(\sqrt{3}-1\right)=20 \\ \Rightarrow h=\frac{20}{\left(\sqrt{3}-1\right)} \\ \Rightarrow h=\frac{20}{\left(\sqrt{3}-1\right)}\times \frac{\left(\sqrt{3}+1\right)}{\left(\sqrt{3}+1\right)} \left[\mathrm{By} \mathrm{rationalization}\right] \\ \Rightarrow h=\frac{20\left(\sqrt{3}+1\right)}{\left(3-1\right)} \\ \Rightarrow h=\frac{20\left(\sqrt{3}+1\right)}{2} \\ \Rightarrow h=10\left(\sqrt{3}+1\right)\mathrm{m} \end{gathered}$$

Hence, the required height of the tower is $10\left(\sqrt{3}+1\right)$ m.

Answer:

Given,
$$\begin{gathered} 1+{\sin }^2\theta =3\sin \theta \cos \theta \\ \Rightarrow \frac{1}{{\sin }^2\theta }+1=\frac{3\cos \theta }{\sin \theta } \left[\mathrm{Dividing} \mathrm{both} \mathrm{sides} \mathrm{by} {\sin }^2\theta \right] \\ \Rightarrow {\mathrm{cosec}}^2\theta +1=3\cot \theta \\ \Rightarrow 1+\mathrm{co}{t}^2\theta +1=3\cot \theta \left[\because {\mathrm{cosec}}^2\theta -{\cot }^2\theta =1\right] \\ \Rightarrow {\cot }^2\theta -3\cot \theta +2=0 \\ \Rightarrow {\cot }^2\theta -2\cot \theta -\cot \theta +2=0 \\ \Rightarrow \cot \theta \left(\cot \theta -2\right)-1\left(\cot \theta -2\right)=0 \\ \Rightarrow \left(\cot \theta -1\right)\left(\cot \theta -2\right)=0 \\ \Rightarrow \cot \theta =1 \mathrm{or} 2 \\ \Rightarrow \tan \theta =1 \mathrm{or} \frac{1}{2} \left[\because \tan \theta =\frac{1}{\cot \theta }\right] \end{gathered}$$

Hence proved.