Prove that: $\;$ $2 \left[\tan^{-1} \left(1\right) + \tan^{-1} \left(\dfrac{1}{2}\right) + \tan^{-1} \left(\dfrac{1}{3}\right)\right] = \pi$
To prove that: $\;$ $2 \left[\tan^{-1} \left(1\right) + \tan^{-1} \left(\dfrac{1}{2}\right) + \tan^{-1} \left(\dfrac{1}{3}\right)\right] = \pi$
i.e. $\;$ To prove that: $\;$ $\tan^{-1} \left(1\right) + \tan^{-1} \left(\dfrac{1}{2}\right) + \tan^{-1} \left(\dfrac{1}{3}\right) = \dfrac{\pi}{2}$
$\begin{aligned}
LHS & = \tan^{-1} \left(1\right) + \tan^{-1} \left(\dfrac{1}{2}\right) + \tan^{-1} \left(\dfrac{1}{3}\right) \\\\
& = \dfrac{\pi}{4} + \tan^{-1} \left(\dfrac{\dfrac{1}{2} + \dfrac{1}{3}}{1 - \dfrac{1}{2} \times \dfrac{1}{3}}\right) \\\\
& \left[\text{Note: } \tan^{-1} A + \tan^{-1} B = \tan^{-1} \left(\dfrac{A + B}{1 - AB}\right)\right] \\\\
& = \dfrac{\pi}{4} + \tan^{-1} \left(\dfrac{5/6}{5/6}\right) \\\\
& = \dfrac{\pi}{4} + \tan^{-1} \left(1\right) \\\\
& = \dfrac{\pi}{4} + \dfrac{\pi}{4} \\\\
& = \dfrac{\pi}{2} = RHS
\end{aligned}$
Hence proved.