Question

SinA-Cos A = 1 /sinA + Cos A prove that

Answer 1

$sinA - cosA = \frac{1}{sinA + cosA}$
Taking RHS

$\frac{1}{sinA + cosA} \\ \\ \frac{1}{sinA + cosA} \times \frac{sinA - cosA}{sinA - cosA} \\ \\ \frac{sinA - cosA}{sin ^{2} A - cos ^{2} A }$
From identity

sin²A – cos²A = 1

$\frac{sinA - cosA}{1} \\ {sinA - cosA}$
SinA-Cos A = LHS

PROVED.