Prove that ${cos}^{2} A - {cos}^{2} B = sin (A + B) sin (B - A)$

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Solution

L.H.S $= {cos}^{2} A - {cos}^{2} B$

$= (cos A + cos B) (cos A - cos B)$

$= (2 cos (\frac{A + B}{2}) cos (\frac{A - B}{2})) (- 2 sin (\frac{A + B}{2}) sin (\frac{A - B}{2}))$

$= (2 sin (\frac{A + B}{2}) cos (\frac{A + B}{2})) (- 2 sin (\frac{A - B}{2}) cos (\frac{A - B}{2}))$

$= - sin (A + B) sin (A - B)$

$= sin (A + B) sin (B - A) =$ R.H.S

Hence proved.

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