Prove trigonometric identity.

Question

How would one go about proving the identity $$ \cot \frac{A}{2} - \tan\frac{A}{2} \equiv 2\cot A $$

I know that $\tan 2A = \dfrac{2\tan A}{1 - \tan^2 A}$ but it is leading me nowhere when implementing it.

Answer 1

$$\cot(A/2) - \tan(A/2) = \frac{1}{\tan (A/2)} - \tan(A/2) = \frac{1-\tan^2(A/2)}{\tan(A/2)}=$$ $$=2\frac{1-\tan^2(A/2)}{2\tan(A/2)}=\frac{2}{\frac{2\tan(A/2)}{1-\tan^2(A/2)}}=\frac{2}{\tan(A)}=2\cot(A)$$

Answer 2

Hint In

$$\cot(A)=\frac{\cos(A)}{\sin(A)}$$

plug

$$\cos(A)=\cos^2(\frac{A}{2})-\sin^2(\frac{A}{2})$$

and

$$\sin(A)=2\sin(\frac{A}{2})\cos(\frac{A}{2}).$$

to get the result.