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#### Question 1:

Choose the correct answer from the given four options:
If , then the value of tan A is
(A) $\frac{3}{5}$

(B) $\frac{3}{4}$

(C) $\frac{4}{3}$

(D) $\frac{5}{3}$

Given: cos A = $\frac{4}{5}$                                                  .....(1)

We have value of cos A, we need to find value of sin A
Also we know that, sin A = √ (1 $-$ cos2 A)           .....(2)                                        (∵ sin2 θ +cos2 θ = 1)
⇒ sin2 A = 1$-$cos2 A
⇒ sin A = √ (1$-$cos2 A)
Thus,
Substituting (1) in (2), we get
Sin A = $\sqrt{\left(1-\frac{16}{25}\right)}$
$\sqrt{\frac{9}{25}}=\frac{3}{5}$
Therefore,
tan A = $\frac{\frac{3}{5}}{\frac{4}{5}}=\frac{3}{4}$
Hence, the correct answer is Option(B).

#### Question 2:

Choose the correct answer from the given four options:
If  then the value of cot A is
(A) $\sqrt{3}$

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

(C) $\frac{\sqrt{3}}{2}$

(D) 1

Given:                                                                       .....(1)
And we know that,                              .....(2)
We need to find the value of cos A.
.....(3)                  (∵ sin2 θ +cos2 θ = 1)
⇒ cos2 A = 1$-$ sin2 A
⇒ cos A = √ (1$-$sin2 A)
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

Hence, the correct answer is Option(A).

#### Question 3:

Choose the correct answer from the given four options:
The value of the expression [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] is
(A) – 1
(B) 0
(C) 1
(D) $\frac{3}{2}$

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).

#### Question 4:

Choose the correct answer from the given four options:
Given that , then cos θ is equal to
(A) $\frac{b}{\sqrt{{b}^{2}-{a}^{2}}}$

(B) $\frac{b}{a}$

(C) $\frac{\sqrt{{b}^{2}-{a}^{2}}}{b}$

(D) $\frac{a}{\sqrt{{b}^{2}-{a}^{2}}}$

Given :
We know that, sin2 θ +cos2 θ =1
⇒ sin2 A = 1 $-$ cos2 A

Hence, the correct answer is Option(C).

#### Question 5:

Choose the correct answer from the given four options:
If cos (α + β) = 0, then sin (α – β) can be reduced to
(A) cos β
(B) cos 2β
(C) sin α
(D) sin 2α

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)                  (∵we have got the value of α, which is 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).

#### Question 6:

Choose the correct answer from the given four options:
The value of (tan1° tan2° tan3° ... tan89°) is
(A) 0

(B) 1

(C) 2

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

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.
= 1
Hence, the correct answer is Option (B).

#### Question 7:

Choose the correct answer from the given four options:
If cos 9α = sinα and 9α< 90°, then the value of tan5α is
(A) $\frac{1}{\sqrt{3}}$

(B) $\sqrt{3}$

(C) 1

(D) 0

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° $-$ $\alpha$)
Thus, sin (90° $-$ 9$\alpha$) = sin$\alpha$                                (∵cos 9$\alpha$ = sin(90° $-$ 9$\alpha$) & 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 ($5×9$) = tan 45° = 1
∴, tan 5$\alpha$ = 1

Hence, the correct answer is Option(C).

#### Question 8:

Choose the correct answer from the given four options:
If ΔABC is right angled at C, then the value of cos (A + B) is

(A) 0

(B) 1

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

(D) $\frac{\sqrt{3}}{2}$

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).

#### Question 9:

Choose the correct answer from the given four options:
If sin A + sin2 A = 1, then the value of the expression (cos2 A + cos4 A) is

(A) 1

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

(C) 2

(D) 3

Given: sin A + sin2 A = 1
⇒ sin A = 1 – sin2 A
(By Rearranging)
⇒ 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).

#### Question 10:

Choose the correct answer from the given four options:
Given that , then the value of (α + β) is
(A) 0°
(B) 30°
(C) 60°
(D) 90°

Given: sin α = 1/2 and cos β = 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).

#### Question 11:

Choose the correct answer from the given four options:
The value of the expression is
(A) 3
(B) 2
(C) 1
(D) 0

Given,

$⇒$

[ ]

=                                                              [  ]

=                                                    [ ]

$\left[\frac{1}{1}+1\right]$
= 2

Hence, the correct answer is Option(B).

#### Question 12:

Choose the correct answer from the given four options:
If is equal to
(A) $\frac{2}{3}$

(B) $\frac{1}{3}$

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

(D) $\frac{3}{4}$

Given: 4 tan θ = 3
$⇒4\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }=3$                                                                  .....(1)

We have to evaluate ,

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

=

$\left(\frac{3-1}{3+1}\right)$ = $\frac{2}{4}$ = $\frac{1}{2}$

Hence, the correct answer is Option(C).

#### Question 13:

Choose the correct answer from the given four options:
If sinθ – cosθ = 0, then the value of (sin4θ + cos4θ) is
(A) 1

(B) $\frac{3}{4}$

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

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

Given: sin θ – cos θ = 0

And we know, tan 45° = 1
So, tan θ = 1 = tan 45°
By comparing above equation, we get θ = 45°

$\left({\left(\frac{1}{\sqrt{2}}\right)}^{4}+{\left(\frac{1}{\sqrt{2}}\right)}^{4}\right)$
$\left(\frac{1}{4}+\frac{1}{4}\right)=\frac{1}{2}$
Hence, the correct answer is Option(C).

#### Question 14:

Choose the correct answer from the given four options:
sin (45° + θ) – cos (45° – θ) is equal to
(A) 2cosθ
(B) 0
(C) 2sinθ
(D) 1

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).

#### Question 15:

Choose the correct answer from the given four options:
A pole 6 m high casts a shadow long on the ground, then the Sun’s elevation is
(A) 60°
(B) 45°
(C) 30°
(D) 90°

Lets 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 = $\frac{BC}{AB}$       (∵ tan θ = $\frac{\mathrm{perpendicular}}{\mathrm{base}}$)

⇒ tan θ = tan 60° (∵ tan 60° = √3)
By comparing, we get
θ = 60°

∴, the sun’s elevation is 60°

Hence, the correct answer is Option(A).

#### Question 1:

Write ‘True’ or ‘False’ and justify your answer in each of the following:

True

(∵ tan (90°$-$ θ) = cot θ)

#### Question 2:

Write ‘True’ or ‘False’ and justify your answer in each of the following:
The value of the expression (cos2 23° – sin2 67°) is positive.

False

cos2 23° $-$ sin2 67° = (cos 23° + sin 67°) (cos 23°$-$ sin 67°)                                  (∵  (a$-$ 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.

#### Question 3:

Write ‘True’ or ‘False’ and justify your answer in each of the following:
The value of the expression (sin 80° – cos 80°) is negative.

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).

#### Question 4:

Write ‘True’ or ‘False’ and justify your answer in each of the following:

LHS =

$\sqrt{\left({\mathrm{sin}}^{2}\theta \right){\mathrm{sec}}^{2}\theta }$                                    (  )

$\sqrt{\frac{\left({\mathrm{sin}}^{2}\theta \right)}{\left({\mathrm{cos}}^{2}\theta \right)}}$
$\frac{\mathrm{sin}\theta }{\mathrm{cos}\theta }$

= RHS

Hence, the given expression is True.

#### Question 5:

Write ‘True’ or ‘False’ and justify your answer in each of the following:
If cos A + cos2A = 1, then sin2A + sin4A = 1.

True

Given: cos A+cos2 A = 1

(Rearranging, and taking cos2 A on the right side of equation)

⇒ cos A = 1$-$ cos2 A                                    (∵ sin2θ  + cos2θ = 1 ⇒ sin2 θ = 1$-$ cos2θ)
⇒ cos A = sin2 A                                           .....(1)
Squaring both sides,
we get cos2 A = sin4 A                                  .....(2)

We have to find whether, sin2A + sin4 A = 1

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

sin2A + sin4 A = cos A + cos2 A                                    (As given)
∴ sin2A+ sin4 A = 1

Hence, the given statement is True.

#### Question 6:

Write ‘True’ or ‘False’ and justify your answer in each of the following:
(tan θ + 2) (2 tan θ + 1) = 5 tan θ + sec2θ.

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                                 (∵ sec2 θ $-$ tan2 θ = 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.

#### Question 7:

Write ‘True’ or ‘False’ and justify your answer in each of the following:
If the length of the shadow of a tower is increasing, then the angle of elevation of the sun is also increasing.

False

Let us understand this by an example $-$ A tower 2√ 3 m high casts a shadow 2 m long on the ground, then the sun’s elevation is 60°

⇒ tan θ = √ 3 = tan 60°
θ = 60°

Figure

Now, if the same height of tower casts a shadow 4m more than that from preceding length, then the Sun’s elevation becomes 30°

θ = 30°

Figure

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.

#### Question 8:

Write ‘True’ or ‘False’ and justify your answer in each of the following:
If a man standing on a platform 3 metres above the surface of a lake observes a cloud and its reflection in the lake, then the angle of elevation of the cloud is equal to the angle of depression of its reflection.

False

Let P be the point where the man is standing and C be the point where the cloud is. MO = 3 m, the platform’s height from the surface of the lake.
The angles, θ= 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 $∆\mathrm{MPC}$,

.....(1)

In $∆\mathrm{PMW}$,

.....(2)

From (1) and (2) we get,

$⇒{\theta }_{1}\ne {\theta }_{2}$

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

#### Question 9:

Write ‘True’ or ‘False’ and justify your answer in each of the following:
The value of 2sinθ can be $a+\frac{1}{a}$, where a is a positive number, and a ≠ 1.

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

By assuming that  $a+\frac{1}{a}$= 2sinθ  is true
Applying the inequality on  , we get
$⇒\frac{a+\frac{1}{a}}{2}\ge \sqrt{\left(a×\frac{1}{a}\right)}\phantom{\rule{0ex}{0ex}}⇒\frac{a+\frac{1}{a}}{2}\ge 1\phantom{\rule{0ex}{0ex}}⇒a+\frac{1}{a}\ge 2$

⇒ sin θ > 1

But $-$1 ≤ sin θ ≤ 1

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

#### Question 10:

Write ‘True’ or ‘False’ and justify your answer in each of the following:
, where a and b are two distinct numbers such that ab > 0.

False

Given: ab 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
, GM = $\sqrt{ab}$
By assuming that    is true statement.

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

So $\frac{{a}^{2}+{b}^{2}}{2}\ge \sqrt{{a}^{2}×{b}^{2}}$

(By AM and GM property as mentioned earlier in the answer)

$⇒\frac{{a}^{2}+{b}^{2}}{2}\ge ab\phantom{\rule{0ex}{0ex}}⇒\frac{{a}^{2}+{b}^{2}}{2ab}\ge 1$
( By our assumption)

But this not possible since,$-$1 ≤ cos θ ≤ 1
Thus, our assumption is wrong

#### Question 11:

Write ‘True’ or ‘False’ and justify your answer in each of the following:
The angle of elevation of the top of a tower is 30°. If the height of the tower is doubled, then the angle of elevation of its top will also be doubled.

False

Let height of tower is h and BC = x m.

Figure

In ∆ABC, tan 30°= AC/BC = h/x

$⇒\frac{1}{\sqrt{3}}=\frac{h}{x}$                                       .....(1)
Now let height be 2h (doubled)
Figure

= 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.

#### Question 12:

Write ‘True’ or ‘False’ and justify your answer in each of the following:
If the height of a tower and the distance of the point of observation from its foot, both, are increased by 10%, then the angle of elevation of its top remains unchanged.

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

$⇒{\theta }_{1}={\mathrm{tan}}^{-1}\left(\frac{h}{x}\right)$                                                               .....(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
$\frac{11}{10}x$

Figure

In ,
.....(2)
$⇒{\theta }_{2}={\mathrm{tan}}^{-1}\left(\frac{h}{x}\right)$
From (1) and (2)  we get θ1 = θ2
Hence, it is true.

#### Question 1:

Prove the following $\frac{\mathrm{sin}\theta }{1+\mathrm{cos}\theta }+\frac{1+\mathrm{cos}\theta }{\mathrm{sin}\theta }=2\mathrm{cosec}\theta$

#### Question 2:

Prove the following

#### Question 3:

Prove the following
If , then

By pythagoras theorem
H2 = P2 + B2 = (3x)2 + (4x)2
H2 = 25x2
∴ H = 5x

Hence proved.

#### Question 4:

Prove the following
(sin α + cos α) (tan α + cot α) = sec α + cosec α.

#### Question 5:

Prove the following

#### Question 6:

Prove the following

#### Question 7:

Prove the following tan θ  + tan (90° – θ) = sec θ sec (90° – θ)

#### Question 8:

Find the angle of elevation of the sun when the shadow of a pole h metres high is metres long.

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

Now in $∆$ ABC,
$\mathrm{tan}\theta =\frac{AB}{BC}=\frac{h}{\sqrt{3}h}=\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}\mathrm{tan}\theta =\frac{1}{\sqrt{3}}\phantom{\rule{0ex}{0ex}}⇒\theta ={\mathrm{tan}}^{-1}\left(\frac{1}{\sqrt{3}}\right)={30}^{o}\phantom{\rule{0ex}{0ex}}⇒\theta ={30}^{o}$

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

#### Question 9:

If  , then find the value of sin2θ – cos2θ .

#### Question 10:

A ladder 15 metres long just reaches the top of a vertical wall. If the ladder makes an angle of 60° with the wall, find the height of the wall.

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

Thus the required height of the wall is 7.5 meters.

#### Question 11:

Simplify (1 + tan2θ) (1 – sinθ) (1 + sinθ)

#### Question 12:

If 2sin2θ – cos2θ = 2, then find the value of θ.

Show that

Given,

#### Question 14:

An observer 1.5 metres tall is 20.5 metres away from a tower 22 metres high. Determine the angle of elevation of the top of the tower from the eye of the observer.

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

Given, AB = 22m , PQ = MB = 1.5m and QB = PM = 20.5m
⇒ AM = AB $-$ MB
⇒ AM = 22 $-$ 1.5 = 20.5m
Now in $∆$APM,
$\mathrm{tan}\theta =\frac{AM}{PM}=\frac{20.5}{20.5}=1\phantom{\rule{0ex}{0ex}}⇒\mathrm{tan}\theta =\mathrm{tan}{45}^{o}\phantom{\rule{0ex}{0ex}}⇒\theta ={45}^{o}$

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

#### Question 15:

Show that tan4θ + tan2θ = sec4θ – sec2θ.

#### Question 1:

If cosecθ + cotθ = p, then prove that $\mathrm{cos}\theta =\frac{{p}^{2}-1}{{p}^{2}+1}$.

Prove that

#### Question 3:

The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20 metres towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower.

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

Now in $∆$PSR

Now in $∆$PQR

Since, after moving 20 m towards the tower the angle of elevation of the top increases by 15o.
i.e.  $\angle$PSR = θ = $\angle$PQR + 15o

#### Question 4:

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

#### Question 5:

Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.

#### Question 6:

The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is $\sqrt{st}$.

Let the height of the tower be h
and $\angle ABC=\theta$
given that BC = s, PC = t
and angle of elevation on both positions are complementary.
i.e ,   ∠APC = 90°$-$θ

So, the required height of the tower is $\sqrt{st}$.

Hence proved.

#### Question 7:

The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. Find the height of the tower.

Let the height of the tower be h and RQ = x m

Given, PR = 50 m
and

Hence, the required height of tower is $25\sqrt{3}$ m

#### Question 8:

A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flag staff are α and β, respectively. Prove that the height of the tower is .

Let the height of the tower be H and OR = x

Given that, height of flag staff = h = FP and ∠PRO = α, ∠FRO = β

Hence proved.

#### Question 9:

If tanθ + secθ = l, then prove that $\mathrm{sec}\theta =\frac{{l}^{2}+1}{2l}$.

Hence proved

#### Question 10:

If sinθ + cosθ = p and secθ + cosecθ = q, then prove that q(p2 – 1) = 2p.

Hence proved

#### Question 11:

If a sinθ + b cosθ = c, then prove that a cosθ – b sinθ = $\sqrt{{a}^{2}+{b}^{2}-{c}^{2}}$.

Hence proved

#### Question 12:

Prove that $\frac{1+\mathrm{sec}\theta -\mathrm{tan}\theta }{1+\mathrm{sec}\theta +\mathrm{tan}\theta }=\frac{1-\mathrm{sin}\theta }{\mathrm{cos}\theta }$

LHS = RHS
Hence proved

#### Question 13:

The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60° and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between the two towers and also the height of the other tower.

Let distance between the two towers = AB = x m
Height of the other tower = PA = h m
Given, height of the tower = QB = 30 m and ∠QAB = 60°, ∠PBA = 30°

Thus the required distance and height is  m respectively.

#### Question 14:

From the top of a tower h m high, the angles of depression of two objects, which are in line with the foot of the tower are α and β (β > α). Find the distance between the two objects.

Let the distance between two objects is x m,
and CD = y m.
Given,
∠BAX = α = ∠ABD,                               [alternate angle]
∠CAY = β = ∠ACD                                  [alternate angle]

Thus, the above equation shows the required distance between two objects.

Hence proved

#### Question 15:

A ladder rests against a vertical wall at an inclination α to the horizontal. Its foot is pulled away from the wall through a distance p so that its upper end slides a distance q down the wall and then the ladder makes an angle β to the horizontal.
Show that $\frac{p}{q}=\frac{\mathrm{cos}\beta -\mathrm{cos}\alpha }{\mathrm{sin}\alpha -\mathrm{sin}\beta }$.

Let OQ = x and OA = y
Given, BQ = q, SA = p, and AB = SQ = Length of ladder
Also, ∠BAO = α and ∠QSO = β

Hence proved

#### Question 16:

The angle of elevation of the top of a vertical tower from a point on the ground is 60°. From another point 10 m vertically above the first, its angle of elevation is 45°. Find the height of the tower.

Let the height of the vertical tower OT = H

and OP = AB = x m
Given, AP = 10 m
TPO = 60o and TAB = 45o

Hence, the required height of the tower is $5\left(3+\sqrt{3}\right)$ m

#### Question 17:

A window of a house is h metres above the ground. From the window, the angles of elevation and depression of the top and the bottom of another house situated on the opposite side of the lane are found to be α and β, respectively. Prove that the height of the other house is h ( 1 + tan α cot β ) metres.

Let the height of the other house = OQ = H
and OB = MW = m
Given, height of first house = WB = h = MO

Thus, the required height of the house is $h\left(1+\mathrm{tan}\alpha .cot\beta \right)$

Hence proved

#### Question 18:

The lower window of a house is at a height of 2 m above the ground and its upper window is 4 m vertically above the lower window. At certain instant the angles of elevation of a balloon from these windows are observed to be 60° and 30°, respectively. Find the height of the balloon above the ground.

Let the height of the balloon from above the ground = H.
and OP = w2R = w1Q = x
Given, the height of the lower window from above the ground  = w2P = 2 m = OR
Height of upper window from above the lower window           = w1w2 = 4 m = QR

Figure

So, the required height is 8 m.

Hence, the required height of the balloon above the ground is 8 m.

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