$\mathrm{Tan} \left(\frac{\mathrm{\pi }}{4}+\frac{1}{2}{\cos }^{-1} \frac{\mathrm{a}}{\mathrm{b}}\right)+\mathrm{Tan}\left(\frac{\mathrm{\pi }}{4}-\frac{1}{2}{\mathrm{Cos}}^{-1} \frac{\mathrm{a}}{\mathrm{b}}\right)$

$$\begin{gathered} put \frac{1}{2}{\mathrm{Cos}}^{-1} \frac{\mathrm{a}}{\mathrm{b}}=\mathrm{\theta }\Rightarrow \mathrm{Cos} 2\mathrm{\theta }=\frac{\mathrm{a}}{\mathrm{b}}; \\ Now \mathrm{Tan} \left(\frac{\mathrm{\pi }}{4}+\frac{1}{2}{\cos }^{-1} \frac{\mathrm{a}}{\mathrm{b}}\right)+\mathrm{Tan}\left(\frac{\mathrm{\pi }}{4}-\frac{1}{2}{\mathrm{Cos}}^{-1} \frac{\mathrm{a}}{\mathrm{b}}\right) \\ =\mathrm{Tan} \left(\frac{\mathrm{\pi }}{4}+\mathrm{\theta }\right)+\mathrm{Tan} \left(\frac{\mathrm{\pi }}{4}-\mathrm{\theta }\right) \end{gathered}$$
$\begin{array}{l}=\frac{1+\tan \theta }{1-\tan \theta }+\frac{1-\tan \theta }{1+\tan \theta }\\ =\frac{(1+\tan \theta {)}^2+(1-\tan \theta {)}^2}{1-{\tan }^2\theta }\\ =\frac{2\left(1+{\tan }^2\theta \right)}{1-{\tan }^2\theta }\\ =\frac{2}{\cos 2\theta }\\ =\frac{2\mathrm{b}}{\mathrm{a}}\end{array}$