Simplify $\cot(2\alpha)-\cot(\alpha)$ without using the double angle formula.

Question

I tried to solve it by the following way: $$\frac{1}{\tan(\alpha+\alpha)}-\frac{1}{\tan(\alpha)}=\frac{-(1+\tan^2(\alpha))}{2\tan(\alpha)}=\frac{2\tan(\alpha)}{\cos^2(\alpha)}.$$ This answer is incorrect, because it must be $$-\frac{1}{\sin(2\alpha)}.$$ How can I solve it?

Answer 1

$$\frac{-(1+\tan^2(\alpha))}{2\tan(\alpha)}=-\frac{1}{2\tan\alpha \cos^2\alpha}=-\frac{1}{2\sin\alpha\cos\alpha}=-\frac{1}{\sin(2\alpha)}.$$

Here we use $$1+\tan^2\alpha=\frac{1}{\cos^2\alpha}.$$

Answer 2

$$\cot2\alpha-\cot\alpha=\frac{\cos2\alpha\sin\alpha-\sin2\alpha\cos\alpha}{\sin2\alpha\sin\alpha}=$$ $$=\frac{-\sin\alpha}{\sin2\alpha\sin\alpha}=-\frac{1}{\sin2\alpha}$$