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Page No 89:

Question 1:

Choose the correct answer from the given four options:
If cos A=45, then the value of tan A is
(A) 35

(B) 34

(C) 43

(D) 53

Answer:

Given: cos A = 45                                                  .....(1)
tan A=sin Acos A
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 = 1-1625
925=35
Therefore, 
tan A = 3545=34
Hence, the correct answer is Option(B).



Page No 90:

Question 2:

Choose the correct answer from the given four options:
If sin A=12 then the value of cot A is
(A) 3

(B) 13

(C) 32

(D) 1

Answer:

Given: sin A=12                                                                      .....(1)
And we know that, cot A =1tan A=cos Asin A                             .....(2)
We need to find the value of cos A.
cos A = 1-sin2A                                                                 .....(3)                  (∵ sin2 θ +cos2 θ = 1)
⇒ cos2 A = 1- sin2 A
⇒ cos A = √ (1-sin2 A)
Substituting (1) in (3), we get
cos A = 1-14 = 34 =32

Substituting values of sin A and cos A in (2), we get
cot A = 3212=3
Hence, the correct answer is Option(A).

Page No 90:

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) 32

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).

Page No 90:

Question 4:

Choose the correct answer from the given four options:
Given that sin θ=ab, then cos θ is equal to
(A) bb2-a2

(B) ba

(C) b2-a2b

(D) ab2-a2

Answer:

Given : sin θ=ab
We know that, sin2 θ +cos2 θ =1
⇒ sin2 A = 1 - cos2 A
sin A = 1-cos2A
cos θ =1-sin2θ=1-ab2= b2-a2b
Hence, the correct answer is Option(C).

Page No 90:

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

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)                  (∵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).

Page No 90:

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) 12

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°.1.1tan 44°.1tan 43°....1tan 3°.1tan 2°.1tan 1°
= 1
Hence, the correct answer is Option (B).



Page No 90:

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) 13

(B) 3

(C) 1

(D) 0

Answer:

Given: cos 9α = sin α and 9α < 90° i.e. 9α 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 9α = sin (90° - α)
Thus, sin (90° - 9α) = sinα                                (∵cos 9α = sin(90° - 9α) & sin(90°- α) = sinα )
⇒ 90°-9αα
⇒ 10α = 90°              (By rearranging)
α = 9°
We have got the value of α i.e. α = 9°
Putting it in tan 5α, we get
tan 5α = tan (5×9) = tan 45° = 1
∴, tan 5α = 1

Hence, the correct answer is Option(C).

Page No 90:

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) 12

(D) 32

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).

Page No 90:

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) 12

(C) 2

(D) 3

Answer:

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).

Page No 90:

Question 10:

Choose the correct answer from the given four options:
Given that sinα=12 and cosβ=12, then the value of (α + β) is
(A) 0°
(B) 30°
(C) 60°
(D) 90°

Answer:

Given: sin α = 1/2 and cos β = 1/2

sin α = 12 = sin 30° and cos β = 12 = 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).    



Page No 91:

Question 11:

Choose the correct answer from the given four options:
The value of the expression sin2 22°+sin2 68°cos2 22°+cos2 68°+sin2 63°+cos 63° sin 27° is
(A) 3
(B) 2
(C) 1
(D) 0

Answer:

Given, sin2 22°+sin2 68°cos2 22°+cos2 68°+sin2 63°+cos 63° sin 27°

 sin2 22°+sin2 68°cos2 22°+cos2 68°+sin2 63°+cos 63° sin (90-63)°                         

sin2 22°+sin2 68°cos2 22°+cos2 68°+sin2 63°+cos 63° cos (63)°                           [sin (90-θ) = cos θ ]

sin2 22°+sin2 68°cos2 22°+cos2 68°+sin2 63°+cos2 63° 

=  sin2 22°+sin2 68°cos2 22°+cos2 68°+1                                                            [ sin2θ + cos2θ = 1 ]

=  sin2 22°+cos2 22°cos2 22°+sin2 22°+1                                                  [sin (90-θ) = cos θ ]

11+1 
= 2 

Hence, the correct answer is Option(B).

Page No 91:

Question 12:

Choose the correct answer from the given four options:
If 4 tanθ=3, then 4 sinθ-cosθ4sinθ+cosθ is equal to
(A) 23

(B) 13

(C) 12

(D) 34

Answer:

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

We have to evaluate ,  4 sinθ-cosθ4sinθ+cosθ

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

 4 sinθ-cosθcosθ4sinθ+cosθcosθ =  4 sinθcosθ-cosθcosθ4 sinθcosθ+cosθcosθ

3-13+1 = 24 = 12

Hence, the correct answer is Option(C).

Page No 91:

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) 34

(C) 12

(D) 14

Answer:

Given: sin θ – cos θ = 0
 sinθ = cosθ
 sinθcosθ = 1
tan θ = 1
And we know, tan 45° = 1
So, tan θ = 1 = tan 45°
By comparing above equation, we get θ = 45°
(sin4θ + cos4θ) = sin4 (45)° + cos4 (45)°
124+124
14+14=12
Hence, the correct answer is Option(C).
 

Page No 91:

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

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).

Page No 91:

Question 15:

Choose the correct answer from the given four options:
A pole 6 m high casts a shadow 23 m long on the ground, then the Sun’s elevation is
(A) 60°
(B) 45°
(C) 30°
(D) 90°

Answer:

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 = BCAB       (∵ tan θ = perpendicularbase)

tanθ = 623

⇒ 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).

 



Page No 93:

Question 1:

Write ‘True’ or ‘False’ and justify your answer in each of the following:
tan 47°cot 43°=1

Answer:

True

tan 47°cot 43°=tan (90-43)°cot 43°=cot 43°cot 43°= 1  (∵ tan (90°- θ) = cot θ)

Page No 93:

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.

Answer:

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.

Page No 93:

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.

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).

Page No 93:

Question 4:

Write ‘True’ or ‘False’ and justify your answer in each of the following:
1-cos2θ sec2θ=tan θ

Answer:

LHS = 1-cos2θ sec2θ

sin2θsec2θ                                    ( sin2θ+cos2θ = 1 )

sin2θ(cos2θ)                                           sec2θ = 1cos2θ
sinθcosθ
tan θ

= RHS

Hence, the given expression is True.

Page No 93:

Question 5:

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

Answer:

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.

Page No 93:

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.

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                                 (∵ 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.

Page No 93:

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.

Answer:

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°
In ACB,tan θ=BCAB=232                             tan θ=PerpendicularBase     

⇒ 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°

In APB,tanθ = ABPB=ABPC+CB 

tanθ=234+2=236=13=tan 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.




Page No 93:

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.

Answer:

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 MPC,

tan θ1=CMPM = hPM
tan θ1h=1PM                                                                    .....(1)


In PMW,

tan θ2=MWPM = MO+OWPM=3+hPM
tan θ23+h=1PM                                                                   .....(2)

From (1) and (2) we get,

tan θ1h=tan θ23+h

θ1θ2

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

Page No 93:

Question 9:

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

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=a+b2  and GM = ab

By assuming that  a+1a= 2sinθ  is true 
Applying the inequality on a and 1a , we get 
a+1a2a×1aa+1a21a+1a2
2 sinθ 2              ( By our assumption)

⇒ sin θ > 1

But -1 ≤ sin θ ≤ 1


∴ Our assumption is wrong and that 2 sin θ cannot be equal to a+1a.


 

Page No 93:

Question 10:

Write ‘True’ or ‘False’ and justify your answer in each of the following:
cos θ=a2+b22ab, where a and b are two distinct numbers such that ab > 0.

Answer:

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
AM = a+b2, GM = ab
By assuming that cos θ=a2+b22ab   is true statement.

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

AM = a2+b22GM = a2b2

So a2+b22a2×b2

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

a2+b22aba2+b22ab1
cos θ 1                                                             ( By our assumption)

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




Page No 93:

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.

Answer:


False

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

Figure

In ∆ABC, tan 30°= AC/BC = h/x

13=hx                                       .....(1)
Now let height be 2h (doubled)
Figure

In  PQR, tanθ = PRQR=2hx

tanθ = 23= 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.






 

Page No 93:

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.

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
Figure
In  ABC 
tan θ1=ACBC=hx
θ1=tan-1hx                                                               .....(1)

Now, when both(height and distance) are increased by ten percent , we get 
New height = h + 10% of  h
1110h

New distance = x + 10% of  x
1110x

Figure

In  PQR,
tan θ2=PRQR=11h1011x10                                                 .....(2)
θ2=tan-1hx
From (1) and (2)  we get θ1 = θ2
Hence, it is true.

 



Page No 95:

Question 1:

Prove the following sinθ1+cosθ+1+cosθsinθ=2cosecθ

Answer:

LHS=sinθ1+cosθ+1+cosθsinθ=sin2θ+1+cosθ2sinθ1+cosθ=sin2θ+1+cos2θ+2cosθsinθ1+cosθ                                            a+b2=a2+b2+2ab=1+1+2cosθsinθ1+cosθ                                                                sin2θ+cos2θ=1=21+cosθsinθ1+cosθ=2sinθ=2cosecθ                              cosecθ=1sinθ=2cosecθ=RHSAs RHS=LHSHence proved

Page No 95:

Question 2:

Prove the following tan A1+sec A-tan A1-sec A=2cosec A

Answer:

LHS=tanA1+secA-tanA1-secA=tanA1-secA-tanA1+secA1+secA1-secA=tanA1-secA-1-secA1-sec2A=tanA-2secA1-sec2A                                                        a+ba-b=a2-b2=2tanA secAsec2A-1  = 2tanA secAtan2A                                                             sec2A-tan2A=1=2 secAtanA=2cosAsinACosA=2sinA=2cosecA=2cosecA=RHSAs, LHS=RHSHence proved                                                  

Page No 95:

Question 3:

Prove the following
If tan A=34, then sin A cos A =1225

Answer:

tanA=34=Perpendicular(P)Base(B)P=3x and B=4x
By pythagoras theorem
H2 = P2 + B2 = (3x)2 + (4x)2
H2 = 25x2
∴ H = 5x



sinA=PH=3x5x=35cosA=BH=4x5x=45Thus, sinA cosA=35×45=1225

Hence proved.


 

Page No 95:

Question 4:

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

Answer:

=sinα+cosαsinαcosα+cosαsinα                                       tanα=sinαcosα and cotα=cosαsinα=sinα+cosαsin2α+cos2αsinαcosα    =sinα+cosα1sinαcosα                                              sin2α+cos2α=1=1cosα+1sinα                                                                   secα=1cosα and cosecα=1sinα=secα+cosecα=RHSAs LHS=RHSHence proved

Page No 95:

Question 5:

Prove the following
3+13-cot 30°=tan3 60°-2 sin 60°

Answer:

RHS=tan360o-2sin60o=33-2×32=33-3=23LHS=3+13-cot30o=3+1×3-3=3+1×3×3-1=332-1=23Thus, LHS=RHSHence proved

Page No 95:

Question 6:

Prove the following 1+cot2 α1+cosec α=cosec α

Answer:

LHS=1+cot2α1+cosecα=1+cos2αsin2α1+1sinα                                         cotα=cosαsinα and cosecα=1sinα=1+cos2αsinα1+sinα=sinα1+sinα+cos2αsinα1+sinα=sinα+sin2α+cos2αsinα1+sinα=1+sinαsinα1+sinα                                          sin2α+cos2α=1=1sinα=cosecα                                         1sinα=cosecα  RHS =cosecαThus, LHS=RHSHence proved

Page No 95:

Question 7:

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

Answer:


LHS=tanθ+tan90-θ=tanθ+cotθ                                          tan90-θ=cotθ=sinθcosθ+cosθsinθ                                    tanθ=sinθcosθ and cotθ=cosθsinθ =sin2θ+cos2θsinθ cosθ                                     sin2θ+cos2θ=1=1sinθcosθ=secθ cosecθ                secθ=1cosθ and cosecθ=1sinθ=secθ sec90-θ                                sec90-θ= cosecθRHS= secθ sec90-θThus, LHS=RHSHence proved

Page No 95:

Question 8:

Find the angle of elevation of the sun when the shadow of a pole h metres high is 3 h metres long.

Answer:

Let the angle of the sun be θ
Given, the height of the pole is h.



Now in  ABC,
tanθ=ABBC=h3h=13tanθ=13θ=tan-113=30oθ=30o

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


 

Page No 95:

Question 9:

If  3 tan θ=1, then find the value of sin2θ – cos2θ .

Answer:

Given, 3tanθ=1tan θ=13θ=tan-113=30oNow, sin2θ-cos2θ=sin230o-cos230o=122-322=14-34=-24=-12sin2θ-cos2θ=-12

Page No 95:

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.

Answer:

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.



In QPR  cos60o=PRPQ=h15cos60o=12=h15h=152=7.5m

Thus the required height of the wall is 7.5 meters.

 

Page No 95:

Question 11:

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

Answer:

1+tan2θ1-sinθ1+sinθ=1+tan2θ1-sin2θ                       a-ba+b=a2-b2=sec2θ.cos2θ                                                                                        1+tan2θ=sec2θ and cos2θ+sin2θ=1=1cos2θ .cos2θ =1                                                                              secθ=1cosθ

Page No 95:

Question 12:

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

Answer:

2sin2θ-cos2θ=22sin2θ-1-sin2θ=2                                   cos2θ+sin2θ=12sin2θ-1+sin2θ=2  3sin2θ=3sin2θ=1sinθ=1θ=sin-11=90o                                            sin90o=1

Page No 95:

Question 13:

Show that cos2 45°+θ+cos2 45°-θtan 60°+θ tan 30°-θ=1

Answer:

Given, 
LHS=cos245o+θ+cos245o-θtan60o+θ.tan30o-θ=cos245o+θ+sin90o-45o-θtan60o+θ.cot90o-30o-θ                              sin90o-θ=cosθ and cot90o-θ=tanθ=cos245o+θ+sin245o+θtan60o+θ.cot60o+θ                                           cos2θ+sin2θ=1=1tan60o+θ.1tan60o+θ =1                                        cotθ=1tanθ1=RHSThus, LHS=RHSHence proved

Page No 95:

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.

Answer:

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,
tanθ=AMPM=20.520.5=1tanθ=tan45oθ=45o

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

Page No 95:

Question 15:

Show that tan4θ + tan2θ = sec4θ – sec2θ.

Answer:

LHS= tan4θ+tan2θ=tan2θtan2θ+1=tan2θ.sec2θ                                          sec2θ=tan2θ+1=sec2θ-1.sec2θ                                tan2θ=sec2θ-1=sec4θ-sec2θ=RHSThus, LHS=RHS



Page No 99:

Question 1:

If cosecθ + cotθ = p, then prove that cosθ=p2-1p2+1.

Answer:

cosecθ+cotθ=p1sinθ+cosθsinθ=p                                          cosecθ=1sinθ and cotθ=cosθsinθ1+cosθsinθ=p1+cosθ2sin2θ=p2                                          Squaring both sides1+2cosθ+cos2θsin2θ=p2Using componendo and dividendo rule1+2cosθ+cos2θ-sin2θ1+2cosθ+cos2θ+sin2θ=p2-1p2+11+2cosθ+cos2θ-1-cos2θ1+2cosθ+cos2θ+sin2θ=p2-1p2+12cosθ+2cos2θ2+2cosθ=p2-1p2+1                                                            cos2θ+sin2θ=1cosθ+cos2θ1+cosθ=p2-1p2+1 cosθ1+cosθ1+cosθ=p2-1p2+1 cosθ=p2-1p2+1 Hence proved

Page No 99:

Question 2:

Prove that sec2θ +cosec2θ=tanθ+cotθ

Answer:

LHS=sec2θ+cosec2θ = 1cos2θ+1sin2θ           secθ=1cosθ and cosecθ=1sinθ=sin2θ+cos2θcos2θsin2θ=1cos2θsin2θ                   sin2θ+cos2θ=1=1cosθsinθ=sin2θcosθsinθ+cos2θcosθsinθ                 sin2θ+cos2θ=1=sinθcosθ+cosθsinθ =tanθ+cotθ =RHSThus, LHS=RHSHence proved

Page No 99:

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.

Answer:

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



Now in PSR
tanθ=PRSR=hxtanθ=hxx=htanθ                                             .....1

Now in PQR
tan30o=PRQR=PRQS+SRtan30o=h20+x20+x=htan30o=h1320+x=h320+htanθ=h3                                                    .....2   From 1
Since, after moving 20 m towards the tower the angle of elevation of the top increases by 15o.
i.e.  PSR = θ = PQR + 15o
θ=30o+15o=45o 20+htan45o=h3                                    From 220+h1=h320=h3-hh3-1=20h=203-1h=203-1×3+13+1                            By rationalizationh=203+13-1=203+12=103+1mHence, the required height of the tower is 103+1m.
 

Page No 99:

Question 4:

If 1 + sin2θ = 3sinθ cosθ, then prove that tanθ = 1 or 12.

Answer:

Given, 1+sin2θ=3sinθ.cosθ1sin2θ+1=3cosθsinθ                                 Dividing both sides by sin2θcosec2θ+1=3cotθ                                1sin2θ=cosec2θ and cosθsinθ=cotθ 1+cot2θ+1=3cotθ                                cosec2θ-cot2θ=1cot2θ-3cotθ+2=0cot2θ-2cotθ-cotθ+2=0cotθcotθ-2-1cotθ-2=0cotθ-1cotθ-2=0cotθ=1 or 2tanθ=1 or 12                                              tanθ=1cotθHence proved

Page No 99:

Question 5:

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

Answer:

sinθ+2cosθ=1sinθ+2cosθ2=1                                         Squaring both sidessin2θ+4cos2θ+4sinθcosθ=1  1-cos2θ+41-sin2θ+4sinθcosθ=1      sin2θ+cos2θ=1-cos2θ-4sin2θ+4sinθcosθ=-4cos2θ+4sin2θ-4sinθcosθ=42sinθ-cosθ2=4                                     a2+b2-2ab=a-b22sinθ-cosθ=2Hence proved

Page No 99:

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 st.

Answer:

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



Now in ABC,   tanθ=ACBC=hs                                        ...1and in APC      tan90o-θ=ACPC            tan90o-θ=cotθcotθ=ht1tanθ=ht                                                     cotθ=1tanθ      ...2On multiplying eq. 1 and 2, we gettanθ.1tanθ=hs.hth2st=1h2=sth=st

So, the required height of the tower is st.

Hence proved.
 

Page No 99:

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.

Answer:

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

Given, PR = 50 m
and SPQ=30o, SRQ=60o



Now in SRQ,      tan60o=SQRQ3=hxh3and in SPQ ,        tan30o=SQPQ=SQPR+RQ=h50+x13=h50+x3.h=50+x3.h=50+h33-13h=503-13h=50 h=5032h=253m

Hence, the required height of tower is 253 m

Page No 99:

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 h tan αtan β-tan α.

Answer:

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



Given that, height of flag staff = h = FP and ∠PRO = α, ∠FRO = β
Now, in PRO,     tanα=PORO=Hx x=Htanα                           ...(1)and in FRO,        tanβ=FORO=FP+POROtanβ=h+Hxx=h+Htanβ                            ...(2)From Eqs 1 and 2,Htanα=h+HtanβHtanβ=h+HtanαHtanβ-Htanα=htanαHtanβ-tanα=htanαH=htanαtanβ-tanαHence, the required height of tower is htanαtanβ-tanα

Hence proved.
 

Page No 99:

Question 9:

If tanθ + secθ = l, then prove that secθ=l2+12l.

Answer:

Given, tanθ+secθ=l                                       ...1    Multiply by secθ-tanθ on numerator and denominator LHS        tanθ+secθsecθ-tanθsecθ-tanθ =l   sec2θ-tan2θsecθ-tanθ =l   1secθ-tanθ =l                            sec2θ-tan2θ=1secθ-tanθ=1l                                            ...2On adding equation 1 and 2, we get2secθ=l+1l  secθ=l2+12l

Hence proved

Page No 99:

Question 10:

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

Answer:

Given, sinθ+cosθ=p                                                        ...1and      secθ+cosecθ=q1cosθ+1sinθ=q                                secθ=1cosθ and cosecθ=1sinθcosθ+sinθcosθsinθ=qpcosθsinθ=q                                          From 1cosθ.sinθ=pq                                                                 ...2cosθ+sinθ=pOn squaring both sides, we getcosθ+sinθ2=p2cos2θ+sin2θ+2sinθcosθ=p2                             a+b2=a2+b2+2ab1+2sinθcosθ=p2                                                  cos2θ+sin2θ=11+2pq=p2   q+2p=p2q 2p=p2q-qqp2-1=2p

Hence proved

Page No 99:

Question 11:

If a sinθ + b cosθ = c, then prove that a cosθ – b sinθ = a2+b2-c2.

Answer:

Given,  asinθ+bcosθ=c asinθ+bcosθ2=c2                                                     squaring both sidesa2sin2θ+b2cos2θ+2ab.sinθcosθ=c2                 x+y2=x2+y2+2xya21-cos2θ+b21-sin2θ+2ab.sinθcosθ=c2 a2-a2cos2θ+b2-b2sin2θ+2ab.sinθcosθ=c2 a2+b2-c2=a2cos2θ+b2sin2θ-2ab.sinθcosθa2+b2-c2=acosθ-bsinθ2                                 x2+y2-2xy=x-y2acosθ-bsinθ2=a2+b2-c2acosθ-bsinθ=a2+b2-c2

Hence proved

Page No 99:

Question 12:

 Prove that 1+secθ-tanθ1+secθ+tanθ=1-sinθcosθ

Answer:

LHS=1+secθ-tanθ1+secθ+tanθ=1+1cosθ-sinθcosθ1+1cosθ+sinθcosθ                                     secθ=1cosθ and tanθ=sinθcosθ=cosθ+1-sinθcosθ+1+sinθ=cosθ+1-sinθcosθ+1+sinθ=2cos2θ2-2sinθ2.cosθ22cos2θ2+2sinθ2.cosθ2     cosθ+1=2cos2θ2 and sinθ=2sinθ2.cosθ2=2cos2θ2-2sinθ2.cosθ22cos2θ2+2sinθ2.cosθ2 =2cosθ2cosθ2-sinθ22cosθ2cosθ2+sinθ2 =cosθ2-sinθ2cosθ2+sinθ2×cosθ2-sinθ2cosθ2-sinθ2                                                    Rationalization=cosθ2-sinθ22cos2θ2-sin2θ2                                                                           a-b2=a2+b2-2ab and a-ba+b=a2-b2cos2θ2+sin2θ2-2cosθ2sinθ2cosθ                                                        cos2θ2-sin2θ2=cosθ=1-sinθcosθ=RHS

LHS = RHS
Hence proved

 

Page No 99:

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.

Answer:

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°


Now in QAB, tan600=QBAB=30x3=30xx=303.33=3033=103mand in PBA,                           tan30o=PAAB=hx13=h103h=10 m
Thus the required distance and height is 103  and 10 m respectively.
 



Page No 100:

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.

Answer:

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



Now in ACD   tanβ=ADCD=hyy=htanβ                                                     ...1and in  ABD  tanα=ADBDtanα=ADBC+CDtanα=hx+yx+y=htanα                                                   ...2y=htanα-xFrom 1 and 2htanβ=htanα-x  x=htanα-htanβ=h1tanα-1tanβ=hcotα-cotβ                               As, cotθ=1tanθ

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

Hence proved

Page No 100:

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 pq=cosβ-cosαsinα-sinβ.

Answer:

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



In, BAOcosα=OAABcosα=yABy=ABcosα=OA                            ...1and  sinα=OBABsinα=OBABOB=BAsinα                                     ...2In, QSOcosβ=OSSQOS=SQcosβ=ABcosβ                                      AB=SQ         ...3and  sinβ=OQSQOQ=SQsinβ=ABsinβ                                         AB=SQ         ...4Now, SA=OS-AOp=ABcosβ-ABcosαp=ABcosβ-cosα                                                                      ...5and BQ=BO=QOq=BAsinα-ABsinβ q=ABsinα-sinβ                                                                          ...6Eq 5 divided by 4, we getpq=ABcosβ-cosαABsinα-sinβ =cosβ-cosαsinα-sinβ pq=cosβ-cosαsinα-sinβ  


Hence proved
 

Page No 100:

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.

Answer:

Let the height of the vertical tower OT = H

and OP = AB = x m
Given, AP = 10 m
TPO = 60o and TAB = 45o  
In, TPO  tan60o=OTOP=Hx3=Hxx=H3                                             ...1In TAB    tan45o=TBAB=H-10x1=H-10xx=H-10H3=H-10                                             From 1H-H3=10H1-13=10 H=1033-1.3+13+1                                        By rationalization=1033+13-1=1033+12=533+1=53+3 m

Hence, the required height of the tower is 53+3 m
 

Page No 100:

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.

Answer:

Let the height of the other house = OQ = H
and OB = MW = m
Given, height of first house = WB = h = MO
and QWM=α, OWM=β=WOB              alternate angleIn, WOB,   tanβ=WBOB=hxx=htanβ                                                            ...1In, QWM,   tanα=QMWM=OQ-MOWMtanα=H-hxx=H-htanα                                                            ...2From 1 and 2htanβ=H-htanαhtanα=H-htanβhtanα=Htanβ-htanβHtanβ=htanα+tanβH=htanα+tanβtanβ=h1+tanαtanβ=h1+tanα.cotβ              As, cotθ=1tanθ

Thus, the required height of the house is h1+tanα.cotβ 

Hence proved

Page No 100:

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.

Answer:

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

           BQ=OB-QR+RO                     =H-4+2                     =H-6and   Bw1Q=30o      Bw2R=60oIn, Bw2R,                tan60o=BRw2R=BQ+QRx 3=H-6+4x x=H-23                                                          ...1In, Bw1Q,               tan30o=BQw1Q               tan30o=H-6x=13 x=3H-6                                                   ...2From 1 and 2       3H-6=H-233H-6=H-23H-18=H-2 2H=16H=8

So, the required height is 8 m.

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



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