Prove $\cot\frac{\theta}{2} - 2\cot\theta = \tan\frac{\theta}{2}$ This is a math exercise I am doing for my year 10 math class. Basically I would like to show how I would prove that $\cot\frac{\theta}{2} - 2\cot\theta = \tan\frac{\theta}{2}$
I was able to expand the LHS and was able to use fractions to get that $$\cot\frac{\theta}{2} - 2\cot\theta = \tan\frac{\theta}{2} = \frac{\cos\frac{\theta}{2} - 2\cos\theta}{\sin\theta}$$
I do not know what I would do from here.
Any help will be appreciated. Thanks :)

suppose $\quad x=\frac \theta 2\quad$so that $\quad \theta=2x$:
$$\cot x-2\cot 2x=\frac {\cos x}{\sin x}-2\frac{\cos 2x}{\sin 2x}=\frac {2\cos^2x-2(\cos^2x-\sin^2x)}{\sin 2x}=\cdots $$
