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