The value of  $2{\tan }^{-1}\left[\sqrt{\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}}\tan \frac{\mathrm{\theta }}{2}\right]$ is equal to

$$\begin{gathered} \Rightarrow 2{\tan }^{-1}\left[\sqrt{\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}}\tan \frac{\mathrm{\theta }}{2}\right]= \\ ={\cos }^{-1}\left(\frac{1-\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}{\tan }^2{\displaystyle \frac{\theta }{2}}}{1+\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}{\tan }^2\frac{\theta }{2}}\right) \\ ={\cos }^{-1}\left(\frac{\left(a+b\right)-(a-b){\tan }^2\frac{\theta }{2}}{\left(a+b\right)+(a-b){\tan }^2\frac{\theta }{2}}\right) \\ ={\cos }^{-1}\left(\frac{a\left(1-{\tan }^2\frac{\theta }{2}\right)+b\left(1+{\tan }^2\frac{\theta }{2}\right)}{a\left(1+{\tan }^2\frac{\theta }{2}\right)+b\left(1-{\tan }^2\frac{\theta }{2}\right)}\right) \\ ={\cos }^{-1}\left(\frac{a{\displaystyle \frac{\left(1-{\tan }^2\frac{\theta }{2}\right)}{\left(1+{\tan }^2\frac{\theta }{2}\right)}}+b}{a+b\frac{\left(1-{\tan }^2\frac{\theta }{2}\right)}{\left(1+{\tan }^2\frac{\theta }{2}\right)}}\right) \\ ={\cos }^{-1}\left[\frac{\mathrm{acos}\theta +\mathrm{b}}{\mathrm{a}+\mathrm{bcos}\mathrm{\theta }}\right] \end{gathered}$$