SSC CHSL 15 Nov 2015 Shift 1 Paper

Question : 149 of 199

Marks: +1, -0

0°<θ<90°

and

cosecθ=c ot ^2θ

, then the value of the expression

cosec ^4θ − 2cosec^ 3θ + cot ^2θ

is equal to:

Solution:

Given :

cosecθ=c ot ^2θ

⇒

cosecθ = cosec^2θ − 1

⇒

cosec^2θ- cosecθ − 1 = 0

------(i)
Squaring both sides, we get :
⇒

cosec^4θ = (cosecθ + 1)^2

⇒

cosec^4θ = cosec^2θ + 2cosecθ+1

-------(ii)
To find :

cosec ^4θ − 2cosec^3θ + cot ^2θ

( cosec^2θ+ 2cosecθ+1)

-2cosec^ 3θ+( cosec^2θ-1)

[Using equation (ii)]
=

−2cosec^ 3θ + 2cosec ^2θ + 2cosecθ

−2cosecθ(cosec ^2θ − cosecθ − 1)

Substituting value from equation (i),
=

−2cosecθ × (0) = 0

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