Question

root over sec square A + cosec square A = sec A cosec A

Answer 1

I hope I helped you. This took a long time to write so, please like and mark as brainliest.

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) 

LCM of cos^2A & sin^2A. which is cos^2A * sin^2A 

now 

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 removing 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.

You're welcome! Have a nice day!