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Question

1sinθ1+sinθ+1+sinθ1sinθ=2cosθ,whereπ2<θπ

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Solution

LHS=1sinθ1+sinθ+1+sinθ1sinθ=(1sinθ)2+(1+sinθ)2(1+sinθ)(1sinθ)=1sinθ+1+sinθ1sin2θ

=2cosθ(1sin2θ=cos2θ1sin2θ=cosθ)

=2cosθ(π2<θ<πcosθ<0)

=RHS Hence Proved.


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