Question

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

Answer 1

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.