Answer
See the explanation below.
Work Step by Step
The expression on the left side $\frac{1-\cos 2x}{\sin 2x}$ can be further simplified by using the double angle formula, that is $\cos 2\theta =1-{{\sin }^{2}}\theta $ and $\sin 2\theta =2\sin \theta \cos \theta $. Therefore, the expression can be written as:
$\begin{align}
& \frac{1-\cos 2x}{\sin 2x}=\frac{1-\left( 1-2{{\sin }^{2}}x \right)}{2\sin x\cos x} \\
& =\frac{1-1+2{{\sin }^{2}}x}{2\sin x\cos x} \\
& =\frac{2{{\sin }^{2}}x}{2\sin x\cos x} \\
& =\frac{\sin x}{\cos x}
\end{align}$
We know the quotient identity $tanx=\frac{\sin x}{\cos x}$. Applying the identity to the expression, we get:
$\frac{\sin x}{\cos x}=\tan x$
Thus, the left side is equal to the right side $\frac{1-\cos 2x}{\sin 2x}=\tan x$.
