Question

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

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

Answer 1

Answer: (B)

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