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) = \(cos^2 A\) – \(sin^2 B\) = \(cos^2 B\) – \(sin^2 A\)

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