Question

$4\,tan^{-1} \frac{1}{5} -tan^{-1} \frac{1}{70} + tan^{-1} \frac{1}{99}$ is equal to

• A. $\pi/6$
• B. $\pi/4$
• C. $\pi/3$
• D. $\pi/2$

Answer 1

Answer: (B)

We have, $4tan^{-1} \frac{1}{5} -tan^{-1} \frac{1}{70} + tan^{-1} \frac{1}{99}$
$= 2\left\{2\,tan^{-1} \frac{1}{5}\right\} - \left\{tan^{-1} \frac{1}{70} -tan^{-1} \frac{1}{99}\right\}$
$= 2\left\{tan^{-1}\left(\frac{2\times1/5 }{1-\left(1/5\right)^{2}}\right)\right\} -tan^{-1}\left\{\frac{\frac{1}{70}-\frac{1}{99}}{1+\frac{1}{70}\times \frac{1}{99}}\right\}$
$= 2\,tan^{-1} \frac{5}{12}\left\{tan^{-1}\left(\frac{29}{6931}\right)\right\}$
$= tan^{-1}\left\{\frac{2\times 5/12 }{1-\left(5/12\right)^{2}}\right\} - tan^{-1}\left(\frac{1}{239}\right)$
$= tan^{-1} \frac{120}{119} - tan^{-1} \frac{1}{239}$
$ = tan^{-1} \left\{\frac{\frac{120}{119} -\frac{1}{239} }{1+\frac{120}{119}\times\frac{1}{239}}\right\}$
$= tan^{-1} 1 = \frac{\pi}{4}$