Prove that:
$\displaystyle 2\arctan\left[\sqrt{\frac{a - b}{a + b}}\tan\frac{x}{2}\right] = \arccos\left[\frac{b + a\cos x}{a + b\cos x}\right]$
We can differentiate each term and show that $\displaystyle f'(x)=g'(x)$ for all $\displaystyle x$.
Thus $\displaystyle f(x)=g(x)+C$ and taking $\displaystyle x=0$ give $\displaystyle C=0$.
There are many calculations but it isn't difficult for all that.