How do you prove$ \frac{1-\cos x}{\sin x}=\tan\left(\frac{x}{2}\right)$?

Question

Now I've seen this before but I forgot how to do it. (by the way, this is without the use of t-formula). It gets difficult when the angles are not the same but I split the fraction on the LHS up and got and stuff in terms of $\csc{x}$ and $\cot{x}$. I then used the identity $1 + cot^2{x} = csc^2{x}$ but am still stuck on reaching something in terms of $\frac{x}{2}$.Is there any neat efficient solution?

Answer 1

$${1-\cos x\over \sin x}={2\sin^2 \frac x2 \over 2\sin \frac x2 \cos \frac x2}=\tan \frac x2$$

Answer 2

You can use the Weierstraß substitution: setting $t=\tan\frac x2$, you have $$\cos x=\frac{1-t^2}{1+t^2},\qquad \sin x =\frac{2t}{1+t^2}.$$