Question

State true or false.
${\tan }^{-1}\left(\frac{1}{3}\right)+{\tan }^{-1}\left(\frac{1}{5}\right)+{\tan }^{-1}\left(\frac{1}{8}\right)+{\tan }^{-1}\left(\frac{1}{7}\right)=\frac{\pi }{2}$

• A. True
• B. False

Answer 1

The correct option is B False

We know that ${\tan }^{-1}x+{\tan }^{-1}y={\tan }^{-1}\left(\frac{x+y}{1-xy}\right),xy<1$

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

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

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

$={\tan }^{-1}\left(\frac{7+3}{21-1}\right)+{\tan }^{-1}\left(\frac{13}{40-1}\right)$

$={\tan }^{-1}\left(\frac{10}{20}\right)+{\tan }^{-1}\left(\frac{13}{39}\right)$

$={\tan }^{-1}\left(\frac{1}{2}\right)+{\tan }^{-1}\left(\frac{1}{3}\right)$

$={\tan }^{-1}\left(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\times \frac{1}{3}}\right)$

$={\tan }^{-1}\left(\frac{5}{5}\right)={\tan }^{-1}1=\frac{\pi }{4}\ne \frac{\pi }{2}$

Hence the given statement is false.