MCQ
$2{\tan ^{ - 1}}\frac{1}{3} + {\tan ^{ - 1}}\frac{1}{2} = $
- A${90^o}$
- B${60^o}$
- C${45^o}$
- ✓${\tan ^{ - 1}}2$
✓
Answer
Correct option: D.
${\tan ^{ - 1}}2$
d
(d) $2\,{\tan ^{ - 1}}\frac{1}{3} + {\tan ^{ - 1}}\frac{1}{2} = {\tan ^{ - 1}}\left( {\frac{{\frac{2}{3}}}{{1 - \frac{1}{9}}}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{2}} \right)$
$ = {\tan ^{ - 1}}\left( {\frac{{\frac{2}{3}}}{{\frac{8}{9}}}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{2}} \right) = {\tan ^{ - 1}}\,\left( {\frac{3}{4}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{2}} \right)$
$ = {\tan ^{ - 1}}\left( {\frac{{\frac{1}{2} + \frac{3}{4}}}{{1 - \frac{1}{2} \times \frac{3}{4}}}} \right) = {\tan ^{ - 1}}(2)$.
(d) $2\,{\tan ^{ - 1}}\frac{1}{3} + {\tan ^{ - 1}}\frac{1}{2} = {\tan ^{ - 1}}\left( {\frac{{\frac{2}{3}}}{{1 - \frac{1}{9}}}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{2}} \right)$
$ = {\tan ^{ - 1}}\left( {\frac{{\frac{2}{3}}}{{\frac{8}{9}}}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{2}} \right) = {\tan ^{ - 1}}\,\left( {\frac{3}{4}} \right) + {\tan ^{ - 1}}\left( {\frac{1}{2}} \right)$
$ = {\tan ^{ - 1}}\left( {\frac{{\frac{1}{2} + \frac{3}{4}}}{{1 - \frac{1}{2} \times \frac{3}{4}}}} \right) = {\tan ^{ - 1}}(2)$.
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