Answer
$\dfrac{\sin2x}{1+\cos2x}=\tan x$
Work Step by Step
$\dfrac{\sin2x}{1+\cos2x}=\tan x$
Substitute $\sin2x$ by $2\sin x\cos x$ and $\cos2x$ by $1-2\sin^{2}x$:
$\dfrac{2\sin x\cos x}{1+(1-2\sin^{2}x)}=\tan x$
$\dfrac{2\sin x\cos x}{2-2\sin^{2}x}=\tan x$
Take out common factor $2$ from the denominator on the left side and simplify:
$\dfrac{2\sin x\cos x}{2(1-\sin^{2}x)}=\tan x$
$\dfrac{\sin x\cos x}{1-\sin^{2}x}=\tan x$
Substitute $1-\sin^{2}x$ by $\cos^{2}x$ and simplify:
$\dfrac{\sin x\cos x}{\cos^{2}x}=\tan x$
$\dfrac{\sin x}{\cos x}=\tan x$
Since $\dfrac{\sin x}{\cos x}=\tan x$, the identity is proved
$\tan x=\tan x$
