Question

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

Answer 1

Formulas :

$1\left)\sin \right(C+D)=2\sin (\frac{C+D}{2}\left)\cos \right(\frac{C-D}{2})$

$2\left)\cos \right(C+D)=2\cos (\frac{C+D}{2}\left)\cos \right(\frac{C-D}{2})$

Rewriting the above equation ;

$\frac{(\sin 9x+\sin 3x)+(\sin 7x+\sin 5x)}{(\cos 9x+\cos 3x)+(\cos 7x+\cos 5x)}$

$\Rightarrow \frac{2\sin \left(6x\right)\cos \left(2x\right)+2\sin \left(6x\right)\cos \left(x\right)}{2\cos \left(6x\right)\cos \left(2x\right)+2\cos \left(6x\right)\cos \left(x\right)}$

$\Rightarrow \frac{\sin 6x}{\cos 6x}=\tan 6x$