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Question

Prove 1sinθ1+sinθ+1+sinθ1sinθ =2cosθ,π2<θ<π.

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Solution

Consider the L.H.S.

=1sinθ1+sinθ+1+sinθ1sinθ

=1sinθ1+sinθ+1+sinθ1sinθ

=1sinθ+1+sinθ1+sinθ1sinθ

=21sin2θ

=2cos2θ

=2cosθ

Since, for π2<θ<π

Therefore,

=2cosθ

=2cosθ

Hence, proved.


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