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Question

# (Sec ^2 A+cosec^2A​​​​​ )square root=tan A+cot A

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Solution

## I solved it step by step, so that u can see what happened in each &every steps The solution of ur question is listed below:- L.H.S = square root of (sec^2 A+cosec^2 A) = square root of (1/cos^2A + 1/sin^2A) = square root of ((sin^2A + cos^2A)/cos^2A * sin^2A) I taken LCMof cos^2A & sin^2A. which is cos^2A * sin^2A now The solution of ur question is listed below:- L.H.S = square root of (1/ cos^2A * sin^2A) because sin^2A + cos^2A =1 so L.H.S = (1/ cosA * sinA) After removed the square root. L.H.S = (sin^2A + cos^2A/ cosA * sinA) I written here (1 = sin^2A + cos^2A) now, L.H.S = (sin^2A/cosA * sinA) + (cos^2A/cosA * sinA) = (sinA/cosA) + (cosA/sinA) = (tanA + cotA) = R.H.S Proved.

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