Prove that: $\;$ $\tan \left[\dfrac{\pi}{4} + \dfrac{1}{2} \cos^{-1} \left(\dfrac{a}{b}\right)\right] + \tan \left[\dfrac{\pi}{4} - \dfrac{1}{2} \cos^{-1} \left(\dfrac{a}{b}\right)\right] = \dfrac{2b}{a}$
Let $\;$ $\cos^{-1} \left(\dfrac{a}{b}\right) = \theta$, $\;\;\;$ $\theta \in \left[0, \pi\right]$
Then, $\;$ $\cos \theta = \dfrac{a}{b}$
$\begin{aligned}
\text{Also, } \tan \left[\dfrac{1}{2} \cos^{-1} \left(\dfrac{a}{b}\right)\right] & = \tan \left[\dfrac{\theta}{2}\right] \\\\
& = \sqrt{\dfrac{1 - \cos \theta}{1 + \cos \theta}} \\\\
& = \sqrt{\dfrac{1 - \dfrac{a}{b}}{1 + \dfrac{a}{b}}} \\\\
& = \sqrt{\dfrac{b - a}{b + a}} \\\\
& = \dfrac{\sqrt{b^2 - a^2}}{b + a} \;\;\; \cdots \; (1)
\end{aligned}$
LHS $= \tan \left[\dfrac{\pi}{4} + \dfrac{1}{2} \cos^{-1} \left(\dfrac{a}{b}\right)\right] + \tan \left[\dfrac{\pi}{4} - \dfrac{1}{2} \cos^{-1} \left(\dfrac{a}{b}\right)\right]$
$= \dfrac{\tan \left(\dfrac{\pi}{4}\right) + \tan \left[\dfrac{1}{2} \cos^{-1} \left(\dfrac{a}{b}\right)\right]}{1 - \tan \left(\dfrac{\pi}{4}\right) \tan \left[\dfrac{1}{2} \cos^{-1} \left(\dfrac{a}{b}\right)\right]} + \dfrac{\tan \left(\dfrac{\pi}{4}\right) - \tan \left[\dfrac{1}{2} \cos^{-1} \left(\dfrac{a}{b}\right)\right]}{1 + \tan \left(\dfrac{\pi}{4}\right) \tan \left[\dfrac{1}{2} \cos^{-1} \left(\dfrac{a}{b}\right)\right]}$
$= \dfrac{1 + \dfrac{\sqrt{b^2 - a^2}}{b + a}}{1 - \dfrac{\sqrt{b^2 - a^2}}{b + a}} + \dfrac{1 - \dfrac{\sqrt{b^2 - a^2}}{b + a}}{1 + \dfrac{\sqrt{b^2 - a^2}}{b + a}}$ $\;\;\; \left[\text{In view of equation } (1)\right]$
$= \dfrac{b + a + \sqrt{b^2 - a^2}}{b + a - \sqrt{b^2 - a^2}} + \dfrac{b + a - \sqrt{b^2 - a^2}}{b + a + \sqrt{b^2 - a^2}}$
$= \dfrac{\left(a + b + \sqrt{b^2 - a^2}\right)^2 + \left(a + b - \sqrt{b^2 - a^2}\right)^2}{\left(a + b + \sqrt{b^2 - a^2}\right) \left(a + b - \sqrt{b^2 - a^2}\right)}$
$= \dfrac{\left(a + b\right)^2 + b^2 - a^2 + 2 \left(a + b\right) \sqrt{b^2 - a^2} + \left(a + b\right)^2 + b^2 - a^2 - 2 \left(a + b\right) \sqrt{b^2 - a^2} }{\left(a + b\right)^2 - \left(b^2 - a^2\right)}$
$= \dfrac{a^2 + b^2 + 2 a b + b^2 - a^2 + a^2 + b^2 + 2 a b + b^2 - a^2}{a^2 + b^2 + 2 a b - b^2 + a^2}$
$= \dfrac{4 b^2 + 4 a b}{2 a^2 + 2 a b}$
$= \dfrac{4 b \left(a + b\right)}{2 a \left(a + b\right)}$
$= \dfrac{2 b}{a} = \text{RHS}$
Hence proved.