Question

${\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\frac{1}{8}$ is equal to

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

Answer 1

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

${\tan }^{-1}\frac{1}{5}+{\tan }^{-1}\frac{1}{7}+{\tan }^{-1}\frac{1}{3}+{\tan }^{-1}\left(\frac{1}{8}\right)$

${\tan }^{-1}\left(\frac{1}{5}\right)+{\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)+{\tan }^{-1}\left(\frac{1}{8}\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{8}{14}\right)+{\tan }^{-1}\left(\frac{15}{55}\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{4}{7}\right)+{\tan }^{-1}\left(\frac{3}{11}\right)$

$\Rightarrow {\tan }^{-1}\left(\frac{44+21}{77-12}\right)={\tan }^{-1}\left(\frac{65}{65}\right)$

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