SSC CHSL 15 Nov 2015 Shift 1 Paper

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Question : 121 of 199

Marks: +1, -0

If

4Sin^2θ-1

and angle θ is less than 90° . Then the value of

Cos^2θ+tan^2θ

is:
Take

(0°<θ<90°)

$12/11$
$13/12$
$17/15$
$11/9$

Solution:

Given :

4Sin^2θ-1=0

⇒

4Sin^2θ=1

⇒

Sin^2θ=1/4

⇒

sinθ=√{1/4}=1/2

⇒

θ = sin^{-1}( 1/2)

⇒

θ = 30°

To find :

Cos^2θ+tan^2θ

=

cos^ 2(30 °) + tan^ 2 (30° )

=

( √3/2 )^2 + ( 1/√3 )^2

=

3/4 + 1/ 3 = 13/12

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