Answer
$2 \sin^2 x \tan (\dfrac{x}{2})=2\sin x - \sin 2x$
Work Step by Step
The half -angle formula can be defined as: $\tan (\dfrac{x}{2})=\dfrac{1- \cos x}{\sin x}$
Now, we have $ 2 \sin^2 x \tan (\dfrac{x}{2})=2 \sin^2 x \times \dfrac{1- \cos x}{\sin x}$
or, $=\dfrac{2 \sin^2 x-2 \sin^2 x \cos x}{\sin x}$
or, $=2 \sin x - 2\sin x \cos x$
or, $= 2\sin x - \sin 2x$
Thus, it has been proven that $2 \sin^2 x \tan (\dfrac{x}{2})=2\sin x - \sin 2x$
