[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.


Buy Best Mathematics E-Books Visit : https://mathstudy.in/

Buy Mathematics Formula Book for Class XI,XII,JEE and other Engineering Competition Exam https://mathstudy.in/product/mathematics-formula-book/

Buy Mathematics Workbook for Class XII ( Fully Solved ) : https://mathstudy.in/product/work-book-class-xii-c-b-s-e-fully-solved/

Buy Mathematics Chapter Tests for Class XII ( Fully Solved) : https://mathstudy.in/product/mathematics-chapter-tests-class-xii-c-b-s-e/

Buy Objective Type Question Bank Class XII (Fully Solved ) : https://mathstudy.in/product/objective-type-question-bank-for-mathematics-class-xii-c-b-s-e/