Solved L.H.S. =cos(90∘−θ)=sinθ∵ L.H.S. = R.H.S.

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Sinθ=Cos(90°-θ), Cosθ=Sin(90°-θ)

Prove that `(cos(90^(@)+theta)sec(-theta)tan(180^(@)-theta))/(sec

Solved Simplify the lefthandside so that LHS=RHS

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