Question

how to solve (sin² A -cos² A)/(sin A -cos A)

Answer 1

As we know that a square -bsquare= (a-b)(a+b) So. (Sina-cosa)(sina +cosa)/sina-cosa =sina+cosa

Answer 2

$\frac{(sin^2A-cos^2A) }{(sinA-cosA)}$ -------------------------------1

$(sin^2A-cos^2A)$ = $(sinA-cosA)(sinA+cosA)$ ----------2

Putting 2 in 1

$\frac{(sinA-cosA)(sinA+cosA) }{(sinA-cosA)}$

= SinA + CosA