Question

Prove that $\frac{(\sin 7x+\sin 5x)+(\sin 9x+\sin 3x)}{(\cos 7x+\cos 5x)+(\cos 9x+\cos 3x)}=\tan 6x$

Answer 1

$L.H.S.=\frac{(\sin 7x+\sin 5x)+(\sin 9x+\sin 3x)}{(\cos 7x+\cos 5x)+(\cos 9x+\cos 3x)}$
$\because \cos A+\cos B=2\cos \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$
and $\sin A+\sin B=2\sin \left(\frac{A+B}{2}\right)\cos \left(\frac{A-B}{2}\right)$
$=\frac{\left[2\sin \left(\frac{7x+5x}{2}\right)\cos \left(\frac{7x-5x}{2}\right)\right]+\left[2\sin \left(\frac{9x+3x}{2}\right)\cos \left(\frac{9x-3x}{2}\right)\right]}{\left[2\cos \left(\frac{7x+5x}{2}\right)\cos \left(\frac{7x-5x}{2}\right)\right]+\left[2\cos \left(\frac{9x+3x}{2}\right)\cos \left(\frac{9x-3x}{2}\right)\right]}=\frac{\left[2\sin \left(\frac{12x}{2}\right)\cos \left(\frac{2x}{2}\right)\right]+\left[2\sin \left(\frac{12x}{2}\right)\cos \left(\frac{6x}{2}\right)\right]}{\left[2\cos \left(\frac{12x}{2}\right)\cos \left(\frac{2x}{2}\right)\right]+\left[2\cos \left(\frac{12x}{2}\right)\cos \left(\frac{6x}{2}\right)\right]}=\frac{2\sin 6x\cos x+2\sin 6x\cos 3x}{2\cos 6x\cos x+2\cos 6x\cos 3x}=\frac{2\sin 6x[\cos x+\cos 3x]}{2\cos 6x[\cos x+\cos 3x]}=\frac{2\sin 6x}{2\cos 6x}=\tan 6x=R.H.S$

Hence proved.