Answer
$\dfrac{\cos 3x-\cos 7x}{\sin 3x+\sin 7x}=\tan x$
Work Step by Step
Need to verify $\dfrac{\cos 3x-\cos 7x}{\sin 3x+\sin 7x}=\tan x$
Consider $\dfrac{\cos 3x-\cos 7x}{\sin 3x+\sin 7x}=\dfrac{-2 \sin (\dfrac{3x+7x}{2})\sin(\dfrac{3x -7x}{2})}{2 \sin (\dfrac{3x+7x}{2})\cos(\dfrac{3x -7x}{2})}$
or, $\dfrac{-2 \sin 5x\sin (-2x)}{2 \sin (5x) \cos (-2x)}=\dfrac{\sin 2x}{\cos 2x}$
Thus,$\dfrac{\sin 2x}{\cos 2x}=\tan x$ (RHS)
Hence, the left-hand side and right hand side are equal.
