$tan \begin{Bmatrix} \frac{\pi}{4}+\frac{1}{2}cos^{-1}\alpha \end{Bmatrix} +tan \begin{Bmatrix}\frac{\pi}{4}-\frac{1}{2}cos^{-1}\alpha \end{Bmatrix}, \alpha ≠ 0 $, is equal to

$\frac{2}{\alpha}$

$tan \begin{Bmatrix} \frac{\pi}{4}+\frac{1}{2}cos^{-1}\alpha \end{Bmatrix} +tan \begin{Bmatrix}\frac{\pi}{4}-\frac{1}{2}cos^{-1}\alpha \end{Bmatrix}$

$= tan \left(\frac{\pi}{4}+\frac{\theta }{2}\right) + tan \left(\frac{\pi}{4}-\frac{\theta }{2}\right), $ where $ \theta = cos^{-1} \alpha $

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