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Question

Prove that :
cosθ1sinθ+cosθ1+sinθ=2secθ

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Solution

We have,
L.H.S. = cosθ1+sinθ+cosθ1+sinθ

L.H.S. = cosθ(1+sinθ)+cosθ(1sinθ)(1sinθ)(1+sinθ)

L.H.S. = cosθ+cosθsinθ+cosθcosθsinθ1sin2θ

L.H.S. = 2cosθcos2θ [1sin2θ=cos2θ]

L.H.S. = 2cosθ=2secθ=R.H.S.

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