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 :)

$\cot \theta=\frac {\cos \theta} {\sin \theta}=\frac {\cos^{2}\frac {\theta} 2 -\sin^{2}\frac {\theta} 2} {2\sin (\frac {\theta} 2) \cos (\frac {\theta} 2)}$. Split this into two terms.
