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Question

Prove that 1+sinθ1sinθ+1sinθ1+sinθ=2secθ

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Solution

LHS=1+sinθ1sinθ+1sinθ1+sinθ

Rationalize the denominators by multiplying suitable factors

=1sinθ1+sinθ×1sinθ1sinθ+1+sinθ1sinθ×1+sinθ1+sinθ

=1sinθcosθ+1+sinθcosθ

=2cosθ

=2secθ

=RHS

Hence proved

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