Question

If $ta{n}^{-1}x=\frac{\pi }{4}-ta{n}^{-1}(\frac{1}{3})$ then $x$ is

• A. $\frac{1}{3}$
• B. $\frac{1}{2}$
• C. $\frac{1}{4}$
• D. $\frac{1}{6}$

Answer 1

Answer: (B)

Given equation ${\tan }^{-1}x=\frac{\pi }{4}-{\tan }^{-1}\left(\frac{1}{3}\right)$
$\Rightarrow {\tan }^{-1}x={\tan }^{-1}(1)-{\tan }^{-1}\left(\frac{1}{3}\right)$
$\left(\because \frac{\pi }{4}={\tan }^{-1}(1)\right)$
$\Rightarrow {\tan }^{-1}x={\tan }^{-1}\left(\frac{1-\frac{1}{3}}{1+\frac{1}{3}}\right)$
$\left[\because {\tan }^{-1}x-{\tan }^{-1}y={\tan }^{-1}\left(\frac{x-y}{1+xy}\right)\right]$
$\Rightarrow {\tan }^{-1}x={\tan }^{-1}\left(\frac{2/3}{4/3}\right)$
$\Rightarrow {\tan }^{-1}x={\tan }^{-1}(1/2)$
$\Rightarrow x=1/2$