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Trigonometric Ratios of Multiples of an Angle

Prove that: ...

Question

Prove that:
$\frac{tan 5 A - tan 3 A}{tan 5 A + tan 3 A} = \frac{sin 2 A}{sin 8 A}$

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Solution

$\frac{tan 5 A - tan 3 A}{tan 5 A + tan 3 A} = \frac{sin 5 A cos 3 A - cos 5 A sin 3 A}{sin 5 A cos 3 A + cos 5 A sin 3 A}$
$= \frac{sin 2 A}{sin 8 A}$

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