Question

Prove that
${\tan }^{-1}\left(\frac{1}{\sqrt{3}}\tan \frac{x}{2}\right)=\frac{1}{2}{\cos }^{-1}\left(\frac{1+2\cos x}{2+\cos x}\right)$

Answer 1

${tan}^{-1}\left(\frac{1}{\sqrt{3}}\right)tan\left(\frac{x}{2}\right)=\left(\frac{1}{2}\right)2{tan}^{-1}\frac{1}{\sqrt{3}}{tan}^{-1}\left(\frac{1}{\sqrt{3}}\right)tan\left(\frac{x}{2}\right)$

$=\left(\frac{1}{2}\right){cos}^{-1}\left[\frac{1-\left(\frac{1}{3}\right){tan}^2\frac{x}{2}}{1+\left(\frac{1}{3}\right){tan}^2\frac{x}{2}}\right]$

$=\left(\frac{1}{2}\right){cos}^{-1}\left[3-tan²\left(x/2\right)/3+tan²\left(x/2\right)\right]\left[\frac{3-tan²\left(\frac{x}{2}\right)}{tan²\left(\frac{x}{2}\right)}\right]$

$=\left(\frac{1}{2}\right){cos}^{-1}\left[\frac{3-(1-cosx)}{\left(\frac{(1+cosx)}{3}\right)}\right]+\left[\frac{(1-cosx)}{(1+cosx)}\right]$

$=\left(\frac{1}{2}\right){cos}^{-1}\left[\frac{(2+4cosx)}{(4+2cosx)}\right]$

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