<s>[INST] 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!
 [/INST] There are several ways to proceed, apart from what Ayman gave
Approach 1: If you know the tangent t-formulas: For $t = \tan \frac {\theta} {2}$, 
$$ \sin \theta = \frac {2t}{1+t^2}, \cos \theta = \frac {1-t^2}{1+t^2}$$
Substitute these into the equation and simplify.
Approach 2: (my preference)If you know that $\tan \frac {\theta}{2} = \frac {\sin \theta}{1+\cos \theta} = \frac { 1- \cos \theta}{\sin \theta}$, then by the equivalence class of fractions, we can sum across the numerator and denominator to get
$$\tan \frac {\theta}{2} = \frac {\sin \theta + 1 - \cos \theta}{ 1 + \cos \theta + \sin \theta}$$
Of course, you could also take the difference across the numerator and denominator to get
$$\tan \frac {\theta}{2} = \frac {\sin \theta - 1 + \cos \theta}{ 1 + \cos \theta - \sin \theta}$$
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