MCQ
$4{\tan ^{ - 1}}\frac{1}{5} - {\tan ^{ - 1}}\frac{1}{{239}}$ is equal to
- A$\pi $
- B$\frac{\pi }{2}$
- C$\frac{\pi }{3}$
- ✓$\frac{\pi }{4}$
$\therefore$ $4{\tan ^{ - 1}}\frac{1}{5} = 2\,\left[ {2{{\tan }^{ - 1}}\frac{1}{5}} \right] = 2{\tan ^{ - 1}}\frac{{\frac{2}{5}}}{{1 - \frac{1}{{25}}}}$
$ = 2{\tan ^{ - 1}}\frac{{10}}{{24}} = {\tan ^{ - 1}}\frac{{\frac{{20}}{{24}}}}{{1 - \frac{{100}}{{576}}}} = {\tan ^{ - 1}}\frac{{120}}{{119}}$
So, $4{\tan ^{ - 1}}\frac{1}{5} - {\tan ^{ - 1}}\frac{1}{{239}} = {\tan ^{ - 1}}\frac{{120}}{{119}} - {\tan ^{ - 1}}\frac{1}{{239}}$
$ = {\tan ^{ - 1}}\frac{{\frac{{120}}{{119}} - \frac{1}{{239}}}}{{1 + \frac{{120}}{{119}}.\frac{1}{{239}}}} = {\tan ^{ - 1}}\frac{{(120 \times 239) - 119}}{{(119 \times 239) + 120}}$
==> ${\tan ^{ - 1}}1 = \frac{\pi }{4}$.
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$\frac{\pi}{2}$
$\frac{1}{2}$
$\frac{\pi}{4}$
$1$
$STATEMENT -1$ : For each real $\mathrm{t}$, there exists a point $\mathrm{c}$ in $[\mathrm{t}, \mathrm{t}+\pi]$ such that $\mathrm{f}^{\prime}(\mathrm{c})=0$. because
$STATEMENT -2$: $f(t)=f(t+2 \pi)$ for each real $t$.