Question

The value of ${\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{8}$ is

• A. $\pi$
• B. $\frac{\pi }{4}$
• C. $\frac{3\pi }{4}$
• D. none of these

Answer 1

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

$ta{n}^{-1}\left(\frac{1}{3}\right)+ta{n}^{-1}\left(\frac{1}{5}\right)+ta{n}^{-1}\left(\frac{1}{7}\right)+ta{n}^{-1}\left(\frac{1}{8}\right)$
$=ta{n}^{-1}\left(\frac{5+3}{15-1}\right)+ta{n}^{-1}\left(\frac{1}{7}\right)+ta{n}^{-1}\left(\frac{1}{8}\right)$
$=ta{n}^{-1}\left(\frac{8}{14}\right)+ta{n}^{-1}\left(\frac{1}{7}\right)+ta{n}^{-1}\left(\frac{1}{8}\right)$
$=ta{n}^{-1}\left(\frac{4}{7}\right)+ta{n}^{-1}\left(\frac{1}{7}\right)+ta{n}^{-1}\left(\frac{1}{8}\right)$
$=ta{n}^{-1}\left(\frac{7(4+1)}{49-4}\right)+ta{n}^{-1}\left(\frac{1}{8}\right)$
$=ta{n}^{-1}\left(\frac{35}{45}\right)+ta{n}^{-1}\left(\frac{1}{8}\right)$
$=ta{n}^{-1}\left(\frac{7}{9}\right)+ta{n}^{-1}\left(\frac{1}{8}\right)$
$=ta{n}^{-1}\left(\frac{9+56}{72-7}\right)$
$=ta{n}^{-1}\left(\frac{65}{65}\right)$
$=ta{n}^{-1}\left(1\right)$
$=\frac{\pi }{4}$