# 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.
Important for :
1
If cos X = m cos Y and where X and Y are not equal to each other and values of angle X and Y range from 0° to 90°, then
» 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.