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

## Solution :

We have,

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

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

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

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

= $$cos^2 A – sin^2 B$$

Now, we can also write it as,

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

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