MCQ
${\tan ^{ - 1}}\frac{3}{4} + {\tan ^{ - 1}}\frac{3}{5} - {\tan ^{ - 1}}\frac{8}{{19}} = $
- ✓$\frac{\pi }{4}$
- B$\frac{\pi }{3}$
- C$\frac{\pi }{6}$
- DNone of these
$ = {\tan ^{ - 1}}\left[ {\frac{{\frac{3}{4} + \frac{3}{5}}}{{1 - \frac{3}{4} \times \frac{3}{5}}}} \right] - {\tan ^{ - 1}}\frac{8}{{19}} = {\tan ^{ - 1}}\frac{{27}}{{11}} - {\tan ^{ - 1}}\frac{8}{{19}}$
$ = {\tan ^{ - 1}}\left[ {\frac{{\frac{{27}}{{11}} - \frac{8}{{19}}}}{{1 + \frac{{27}}{{11}} \times \frac{8}{{19}}}}} \right] = {\tan ^{ - 1}}\left( {\frac{{425}}{{425}}} \right) = {\tan ^{ - 1}}(1) = \frac{\pi }{4}$.
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