Question

Prove that$\frac{\cos (7x-5y)+\cos (7y-5x)}{\sin (7x-5y)+\sin (7y-5x)}=\cot (x+y)$

Answer 1

L.H.S

$\Rightarrow \frac{\cos (7x-5y)+\cos (7y-5x)}{\sin (7x-5y)+\sin (7y-5x)}$

We know that

$\cos C+\cos D=2\cos \frac{C+D}{2}\cdot \cos \frac{C-D}{2}$

$\sin C+\sin D=2\sin \frac{C+D}{2}\cdot \cos \frac{C-D}{2}$

Therefore,

$\Rightarrow \frac{2\cos \frac{7x-5y+7y-5x}{2}\cdot \cos \frac{7x-5y-7y+5x}{2}}{2\sin \frac{7x-5y+7y-5x}{2}\cdot \cos \frac{7x-5y-7y+5x}{2}}$

$\Rightarrow \frac{2\cos \frac{2x+2y}{2}}{2\sin \frac{2x+2y}{2}}$

$\Rightarrow \frac{\cos (x+y)}{\sin (x+y)}$

$\Rightarrow \cot (x+y)$

R.H.S

Hence, proved.