Question

(1 cotA)/secA = sin^2A/(1-cosA)

Answer 1

Taking RHS

(1 cotA)/secA = sin^2A/(1-cosA)

Sin A ^ 2 = 1 – cos A ^2

Hence, 1 – cos A ^2//(1-cosA)

A^2 – B^2 = (A – B) (A + B)

(1 - cosA) (1 + cosA) /(1-cosA)

(1+ cosA)

Cos A = 1/sec A

1 + 1/secA

secA + 1/secA = RHS

the LHS part of the question doesn’t make sense.