Question

Prove that sinA +2sin3A /2cos3A +cosA=tanA

Answer 1

There is a mistake in the given problem. The coefficeint of cosA in the denominator must be 5.   or, the coefficient of sinA in the numerator could be someother number.

$LHS=\frac{sinA +2sin3A}{2cos3A +5cosA}\\\\=\frac{sinA+2(3sinA-4sin^3A)}{2(4cos^3A-3cosA)+5cosA}\\\\=\frac{SinA}{CosA}*\frac{7-8\ Sin^2A}{8\ cos^2A-1}\\\\=tanA\frac{7-8sin^2A}{8-8sin^2A-1}\\\\=tanA$