prove that $\frac{\sin 3 A-\sin A}{\cos 3 A+\cos A}=\tan A$

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written 5.6 years ago by

teamques10 ★ 69k

Solution:

$\mathrm{LHS}=\frac{\sin 3 A-\sin A}{\cos 3 A+\cos A}$

$=\frac{2 . \cos \left(\frac{3 A+A}{2}\right) \cdot \sin \left(\frac{3 A-A}{2}\right)}{2 . \cos \left(\frac{3 A+A}{2}\right) \cdot \cos \left(\frac{3 A-A}{2}\right)}$

$=\frac{2 \cos 2 A \sin A}{2 \cos 2 A \cdot \cos A}$

$=\tan A$

$=\mathrm{RHS}$

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