Prove that sin2A-sin2Bsin A cos A-sin B cos B-tan A-B

Let us consider the problem sin^2A sin^2Bsin Acos A sin Bcos B = 1 cos 2A2 1 cos 2B22sin Acos A2 2 sinB cosB2 =1 cos 2A 1 + cos 2Bsin 2 ...

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Standard XII

Mathematics

Trigonometric Ratios Using Right Angled Triangle

Prove that ...

Question

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

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Solution

Let us consider the problem
$\frac{s i n^{2} A - s i n^{2} B}{sin A cos A - sin B cos B} = \frac{\frac{1 - cos 2 A}{2} - \frac{1 - cos 2 B}{2}}{\frac{2 sin A cos A}{2} - \frac{2 sinB cosB}{2}}$
$= \frac{1 - cos 2 A - 1 + cos 2 B}{sin 2 A - sin 2 B}$

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

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

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