SSC CHSL Model Paper 12

Question : 73 of 100

Marks: +1, -0

If sinθ + cosecθ = 2 then sin 2θ + cosec 2θ is

Solution:

We have, (sinθ + cosec θ) = 2
Squaring both sides, we get:

(\sin θ + \cosec θ)^2 = 2^2

⇒

\sin^2θ + \cosec^2θ + 2 × \sin θ

× \cosec θ = 4

⇒

\sin^2θ + \cosec^2θ + 2 × \sin θ ×1/{\sinθ}

⇒

\sin^2θ + \cosec^2θ+2=4

⇒

\sin^2θ + \cosec^2θ=2

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