Question

Prove that:
$\frac{(sin7x+sin5x)+(sin9x+sin3x)}{(cos7x+cos5x)+(cos9x+cos3x)}=tan6x$

Answer 1

$LHS=\frac{(sin7x+sin5x)+(sin9x+sin3x)}{(cos7x+cos5x)+(cos9x+cos3x)}$

$=\frac{2\sin \left(\frac{7x+5x}{2}\right)\cos \left(\frac{7x-5x}{2}\right)+2\sin \left(\frac{9x+3x}{2}\right)\cos \left(\frac{9x-3x}{2}\right)}{2\cos \left(\frac{7x+5x}{2}\right)\cos \left(\frac{7x-5x}{2}\right)+2\cos \left(\frac{9x+3x}{2}\right)\cos \left(\frac{9x-3x}{2}\right)}$

$=\frac{\sin 6x\cos x+\sin 6x\cos 3x}{\cos 6x\cos x+\cos 6x\cos 3x}=tan6x=RHS$

Hence proved