# Prove that sin (A + B) sin (A – B) = $$sin^2 A$$ – $$sin^2 B$$.

## Solution :

We have,

sin (A + B) sin (A – B) = (sin A cos B + cos A sin B) (sin A  cos B – cos A sin B)

= $$sin^2 A cos^2 B$$ – $$cos^2 A sin^2 B$$

= $$sin^2 A (1 – sin^2 B)$$ – $$(1 – sin^2 A) sin^2 B$$

= $$sin^2 A$$ – $$sin^2 A sin^2 B$$ – $$sin^2 B$$ + $$sin^2 A sin^2 B$$

= $$sin^2 A – sin^2 B$$

Now, we can also write it as,

= $$(1 – cos^2 A)$$ – $$(1 – cos^2 B)$$ = $$cos^2 B$$ – $$cos^2 A$$

Hence, sin (A + B) sin (A – B) = $$sin^2 A$$ – $$sin^2 B$$ = $$cos^2 B$$ – $$cos^2 A$$