Show that $2{\tan ^{ - 1}}( - 3) = \frac{{ - \pi }}{2} + {\tan ^{ - 1}}\left( {\frac{{ - 4}}{3}} \right)$.

${\rm{LHS}} = 2{\tan ^{ - 1}}( - 3) = - 2{\tan ^{ - 1}}3$

$= - \left[ {{{\cos }^{ - 1}}\frac{{1 - {3^2}}}{{1 + {3^2}}}} \right]$

$= - \left[ {{{\cos }^{ - 1}}\left( {\frac{{ - 8}}{{10}}} \right)} \right] = - \left[ {{{\cos }^{ - 1}}\left( {\frac{{ - 4}}{5}} \right)} \right]$

$= - \left[ {\pi - {{\cos }^{ - 1}}\left( {\frac{4}{5}} \right)} \right]$

$= - \pi + {\cos ^{ - 1}}\left( {\frac{4}{5}} \right)$

$\left[ {{\rm{let}}{{\cos }^{ - 1}}\left( {\frac{4}{5}} \right) = \theta \Rightarrow \cos \theta = \frac{4}{5} \Rightarrow \tan \theta = \frac{3}{4} \Rightarrow \theta = {{\tan }^{ - 1}}\frac{3}{4}} \right]$

$= - \pi + {\tan ^{ - 1}}\left( {\frac{3}{4}} \right) = - \pi + \left[ {\frac{\pi }{2} - {{\cot }^{ - 1}}\left( {\frac{3}{4}} \right)} \right]$

$= - \frac{\pi }{2} - {\cot ^{ - 1}}\frac{3}{4} = - \frac{\pi }{2} - {\tan ^{ - 1}}\frac{4}{3}$

$= - \frac{\pi }{2} + {\tan ^{ - 1}}\left( {\frac{{ - 4}}{3}} \right)$

= RHS $\quad$ Hence proved.