Question

Prove that $2{\tan }^{-1}\left(\sqrt{\frac{a-b}{a+b}}\tan \frac{x}{2}\right)={\cos }^{-1}\left(\frac{a\cos x+b}{a+b\cos x}\right)$.

Answer 1

LHS $=2{\tan }^{-1}[\frac{\sqrt{a-b}}{\sqrt{a+b}}.\frac{sinx/2}{cosx/2}]$

$={\cos }^{-1}\left[\frac{1-\frac{a-b}{a+b}\frac{{\sin }^{2}x/2}{{\cos }^{2}x/2}}{1+\frac{a-b}{a+b}\frac{{\sin }^{2}x/2}{{\cos }^{2}x/2}}\right]$

$(\because {2\tan }^{-1}x={\cos }^{-1}\left(\frac{1-x^2}{1+x^2}\right))$

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

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

Using $\cos 2\theta ={\cos }^2\theta -{\sin }^2\theta$

$\Rightarrow cosx={\cos }^{2}x/2-{\sin }^{2}x/2$

$={\cos }^{-1}\left[\frac{acosx+b}{a+bcosx}\right]$

$=RHS$