SSC CHSL 6 Dec 2015 Shift 2 Paper

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Question : 127 of 200

Marks: +1, -0

If

{Sinθ+Cosθ}/{Sinθ-Cosθ}=3

then the value

Sin^4θ

of is:

$4/5$
$2/5$
$1/5$
$3/5$

Solution:

Expression :

{Sinθ+Cosθ}/{Sinθ-Cosθ}=3

⇒

sinθ + cosθ

= 3sinθ − 3cosθ

⇒

3sinθ − sinθ

= cosθ + 3cosθ

⇒

2sinθ = 4cosθ

⇒

sinθ = 2 √{1 − sin^2 θ}

Squaring both sides, we get :
⇒

sin ^2θ = 4(1 − sin^2 θ)

⇒

sin^ 2θ = 4 − 4sin^2 θ

⇒

sin^2θ + 4sin^2θ = 4

⇒

5sin^2θ = 4

⇒

sin^2θ =4/ 5

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