$\begin{aligned} & 2 \tan ^{-1}\left[\sqrt{\frac{a-b}{a+b}} \tan \frac{\theta}{2}\right]=\cos ^{-1}\left[\frac{1-\left(\frac{a-b}{a+b}\right) \tan ^2 \frac{\theta}{2}}{1+\left(\frac{a-b}{a+b}\right) \tan ^2 \frac{\theta}{2}}\right] \\ & \left(2 \tan ^{-1} x=\cos ^{-1} \frac{1-x^2}{1+x^2}\right) \\ & =\cos ^{-1}\left[\frac{(a+b)-(a-b) \tan ^2 \frac{\theta}{2}}{(a+b)+(a-b) \tan ^2 \frac{\theta}{2}}\right] \\ & =\cos ^{-1}\left[\frac{a\left(1-\tan ^2 \frac{\theta}{2}\right)+b\left(1+\tan ^2 \frac{\theta}{2}\right)}{a\left(1+\tan ^2 \frac{\theta}{2}\right)+b\left(1-\tan ^2 \frac{\theta}{2}\right)}\right] \\ & =\cos ^{-1}\left[\frac{\frac{a\left(1-\tan ^2 \frac{\theta}{2}\right)}{1+\tan ^2 \frac{\theta}{2}}+b}{a+b\left(\frac{1-\tan ^2 \frac{\theta}{2}}{1+\tan ^2 \frac{\theta}{2}}\right)}\right]=\cos ^{-1}\left[\frac{a \cos \theta+b}{a+b \cos \theta}\right] \\ & \end{aligned}$