Question

$4{\tan }^{-1}\frac{1}{5}-{\tan }^{-1}\frac{1}{239}$ is equal to

• A. $\pi$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{3}$
• D. $\frac{\pi }{4}$

Answer 1

The correct option is D $\frac{\pi }{4}$

We have,

$4{\tan }^{-1}\frac{1}{5}$

$=2\left(2{\tan }^{-1}\frac{1}{5}\right)$

$=2{\tan }^{-1}\left(\frac{2\left(\frac{1}{5}\right)}{1-{\left(\frac{1}{5}\right)}^2}\right)$

$=2{\tan }^{-1}\left(\frac{2\times 5}{25-1}\right)$

$=2{\tan }^{-1}\frac{10}{24}$

$={\tan }^{-1}\left(\frac{2\left(\frac{10}{24}\right)}{1-{\left(\frac{10}{24}\right)}^2}\right)$
$={\tan }^{-1}\left(\frac{20\times 24}{576-100}\right)$

$={\tan }^{-1}\frac{120}{119}$

Therefore,

$4{\tan }^{-1}\frac{1}{5}-{\tan }^{-1}\frac{1}{239}$

$={\tan }^{-1}\frac{120}{119}-\frac{1}{239}$

$={\tan }^{-1}\left(\frac{\frac{120}{119}-\frac{1}{239}}{1+\left(\frac{120}{119}\right)\left(\frac{1}{239}\right)}\right)$

$\approx {\tan }^{-1}\left(1\right)=\frac{\pi }{4}$

Hence, this is the answer.