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

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Sin 90-theta by 1+sin theta +cos theta by 1-cos 90 theta =2sec theta

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Is LHS {sin (180+A) ×cos (90+A) ×tan (270+A)} / {sin (540-A) ×cos (270+A) × cos (360+A)} equal to RHS(1) -cosecA or RHS(2) +cosecA? - Quora

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