Proof of tangent half identity 
Prove the following: $$\tan \left(\frac{x}{2}\right) = \frac{1 + \sin (x) - \cos (x)}{1 + \sin (x) + \cos (x)}$$

I was unable to find any proofs of the above formula online. Thanks!

You could always use Euler's formula, setting $$ z = e^{i x} $$ which gives for the left
$$ \tan\left(\frac{x}{2}\right) = \frac{1}{i} \frac{z^{1/2}-z^{-1/2}}{z^{1/2}+z^{-1/2}}
= \frac{1}{i} \frac{z-1}{z+1}$$
and for the right
$$ \frac{1+\sin(x)-\cos(x)}{1+\sin(x)+\cos(x)} =
\frac{1+ 1/(2i) z - 1/(2i) 1/z - (1/2) z - (1/2) 1/z}
{1+ 1/(2i) z - 1/(2i) 1/z + (1/2) z + (1/2) 1/z} =
\frac{z + 1/(2i) z^2 - 1/(2i) - (1/2) z^2 - (1/2)}
{z + 1/(2i) z^2 - 1/(2i) + (1/2) z^2 + (1/2)} $$
which is
$$ \frac{-1/2(1+i)(z-1)(z+i)}{1/2(1-i)(z+1)(z+i)} =
-\frac{1+i}{1-i}  \frac{z-1}{z+1} = \frac{1}{i} \frac{z-1}{z+1}.$$
While this may not be the most intuitive approach it does show that these trigonometric identities are amenable to automated theorem proving -- just make the substitution from Euler's formula, factorize LHS and RHS, done.
