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# Prove that LHS=RHS:(1/sec^2A-cos^2A+1/cosec^2A-sin^2A)sin^2Acos^2A=1-sin^2Acos^2A/2+sin^2Acos^2A?

Plz answer this quest....its urgent!!!

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- SivaLv 51 decade agoFavorite Answer
(1/sec^2A-cos^2A+1/cosec^2A-sin^2A)sin^2Acos^2A=1-sin^2Acos^2A/2+sin^2Acos^2A

let us take the LHS first

(1/sec^2A-cos^2A+1/cosec^2A-sin^2A)sin^2Acos^2A

({1/sec^2A}-cos^2A+{1/cosec^2A}-sin^2A)sin^2Acos^2A({cos^2A}-cos^2A+{sin^2A}-sin^2A)sin^2Acos^2A

(cos^2A-cos^2A+sin^2A-sin^2A)sin^2Acos^2A

(0)sin^2Acos^2A

0

now let us take the RHS

1-sin^2Acos^2A/2+sin^2Acos^2A

(1-sin^2Acos^2A)/(2+sin^2Acos^2A)

i'm not sure how to solve the RHS

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