Question

root of sec2A+cosec2A=tanA+cotA

Answer 1

Root sec2A + cosec2A
=root 1/cos2A +1/sin2A
=root sin2A+cos2A / sin2Acos2A
=root 1/ sin2Acos2A
=1/sinAcosA
=sin2A+cos2A/sinA cosa (1=sin2A+cos2A)
=tan A + cot A
Hope this helps...... Mark it brainliest pls

Answer 2

Answer:

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.