vii) sin theta/1+cos theta + 1+cos theta/sin theta = 2 cosec theta
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Answer:
LHS=[sinθ/(1+cosθ)]+[(1+cosθ)/sinθ]
=[sin²θ+(1+cosθ)²]/[sinθ(1+cosθ)]
=(sin²θ+1+2cosθ+cos²θ)/[sinθ(1+cosθ)]
=(2+2cosθ)/[sinθ(1+cosθ)]
=[2(1+cosθ)]/[sinθ(1+cosθ]
=2/sinθ
=2cosecθ=RHS
∴LHS=RHS proved
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