[NCERT,Exemplar.2.3,Q.6,Page.35]

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

# [NCERT,Exemplar.2.3,Q.6,Page.35]

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

$ = – \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.

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