# Proof of $\cos^{-1}[\frac{\cos(a) + \cos(b)}{1 + \cos(a)\cos(b)}] = 2\tan^{-1}[\tan(\frac{a}{2})\tan(\frac{b}{2})]$ by integration

Is it possible to prove

$$\cos^{-1}\left[\frac{\cos(a) + \cos(b)}{1 + \cos(a)\cos(b)}\right] = 2\tan^{-1}\left[\tan\left(\frac{a}{2}\right)\tan\left(\frac{b}{2}\right)\right]$$

with the help of integration?

I know how to prove it without integration but can't develop any approach to solve it using integration.

• I think you cant as integration operator would just get cancelled a d if integration of someone gives another then they arent equal as its area under the curve for the integral function – Archis Welankar Feb 25 '16 at 6:59
• Are there any restrictions on $a$ and $b$? The left side is always positive, $range(\arccos)=[0,\pi]$, but if $a·b<0$ then the right side is negative. – LutzL Feb 25 '16 at 11:25

If $b=0$ then the equation reduces to $$\arccos(1)=2\arctan(0)$$ which is true. Then compute the derivatives of both sides for fixed $a$ and variable $b$ and hopefully both will be the same.
In other news, $$\arccos x=2\arctan y \iff y=\tan(\tfrac12\arccos x)=\frac{\sin\arccos x}{1+\cos\arccos x}=\frac{\sqrt{1-x^2}}{1+x}=\sqrt{\frac{1-x}{1+x}}$$ or $$x=\frac{1-y^2}{1+y^2}$$