Question

$sinA-2 sin 3A/2cos 3 A-cos A=tan A$

Answer 1

The answer is given below :

Now, L.H.S.

$= \frac{sin \alpha - 2 {sin}^{3} \alpha }{2 {cos}^{3} \alpha - cos \alpha } \\ \\ = \frac{sin \alpha (1 - 2 {sin}^{2} \alpha ) }{cos \alpha (2 {cos}^{2} \alpha - 1) } \\ \\ = \frac{sin \alpha }{cos \alpha } \times \frac{cos2 \alpha }{cos2 \alpha } \\ \\ = \frac{sin \alpha }{cos \alpha } \\ \\ = tan \alpha$

= R.H.S. [Proved]

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