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

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