Trigonometry Question and Answer for SSC CGL Tier 1, CGL Tier 2, Railways, CDSE, 2018

Directions : Read the following questions carefully and choose the right answer.
» Explain it
A
cos X = m cos Y

cos X  =  m
cos Y 1

By componendo and dividendo,
cos X + cos Y  =  m + 1
cos X – cos Y m – 1

2 cos ( X + Y ) . cos ( X – Y )
2 2
 = 
m + 1
m – 1
– 2 Sin ( X + Y ) . sin ( X – Y )
2 2


cos ( X + Y )
2
.
cos ( Y – X )
2
 = 
m + 1
m – 1
sin ( X + Y )
2
sin ( Y – X )
2

∵ cos (-X) = cos X whereas sin (- X) = - sin X

cot X + Y . cot Y – X  =  m + 1
2 2 m – 1

cot X + Y  =  m + 1 . tan Y – X
2 m – 1 2

[ ∵  cot Θ =  1 ]
tan Θ

Hence, option A is correct.

2
Evaluate : ( CotΘ – CosecΘ + CotΘ + CosecΘ 
» Explain it
B
Given, ( CotΘ – CosecΘ + CotΘ + CosecΘ ) 

= (CotΘ + CotΘ) – ( CosecΘ – CosecΘ)

=  CotΘ ( CotΘ + 1)  – CosecΘ ( CosecΘ  – 1 )
                                                                                                  
=  CotΘ.Cosec2Θ  – CosecΘ.CotΘ                             [∵  ( CotΘ + 1) = Cosec2Θ & ( CosecΘ  – 1 ) = CotΘ]

= 0

Hence, option B is correct.

3
 If 7 sin2 Θ + 3 cos2 Θ = 4 and 0 ≤ Θ ≤  π  , then the value of tan Θ is  :
2
» Explain it
C
7 sin2 Θ + 3 cos2= 4

⇒ 7 sin2 Θ + 3(1 – sin2 Θ )= 4 ⇒ sin2 Θ  =  1  .so, sin Θ =  1
4 2

  cos Θ = 
1 – sin2 Θ
 = 
1  –  1
4
 = 
3
2

  tan Θ =  sin Θ  = ( 1  ×  2 ) 1
cos Θ 2
3
3

Hence, option C is correct.

4
If cosΘ + secΘ = 2 ,then the value of  cos68Θ  +  sec68Θ  equal to
» Explain it
B
cosΘ + secΘ = 2

or, cosΘ +  1  = 2
cosΘ

or, cos2Θ + 1 = 2 cosΘ

or, cos2Θ – 2cosΘ + 1 = 0

or, (cosΘ – 1)= 0

cosΘ = 1

∴  secΘ =  1  = 1
cosΘ

Now, cos68Θ + sec68Θ = 1 + 1 = 2

Hence, option B is correct.

5
If x = a cos3 Θ, y = b sin3 Θ, then

( x ) 2/3  +  ( y ) 2/3  = ?
a   b  
» Explain it
A
x = a cos3 Θ

∴   x  = cos3 Θ
a

and y = b sin3 Θ
∴   y  = sin3 Θ
b

Now,  ( x ) 2/3  +  ( y ) 2/3
a   b  

= (cos3 Θ)2/3 + (sin3 Θ)2/3

= cos2 Θ + sin2 Θ = 1                      [  cos2 Θ + sin2 Θ = 1]

Hence, option A is correct.