MCQ
$2{\tan ^{ - 1}}5 + {\tan ^{ - 1}}\frac{5}{{12}} = ......$
- A$\frac{\pi }{4}$
- B$\frac{{2\pi }}{3}$
- ✓$\pi $
- D$\frac{\pi }{2}$
$2{\tan ^{ - 1}}5 + {\tan ^{ - 1}}\frac{5}{{12}}$
$ = \pi + {\tan ^{ - 1}}\left( {\frac{{2\left( 5 \right)}}{{1 - 25}}} \right) + {\tan ^{ - 1}}\frac{5}{{12}}$
$\left(\because{2{{\tan }^{ - 1}}x = {{\tan }^{ - 1}}\frac{{2x}}{{1 - {x^2}}} + \pi ,} where , x>1\right)$
$ = \pi + {\tan ^{ - 1}}\left( { - \frac{{10}}{{24}}} \right) + {\tan ^{ - 1}}\frac{5}{{12}}$
$ = \pi - {\tan ^{ - 1}}\frac{5}{{12}} + {\tan ^{ - 1}}\frac{5}{{12}} = \pi $
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