Q96 / 180 · KCET 2011 · Mathematics
If $\tan ^{-1} x=\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{3}\right)$, then $x$ is
- A$\frac{1}{3}$
- Correct answerB$\frac{1}{2}$
- C$\frac{1}{4}$
- D$\frac{1}{6}$
Solution
Given equation $\tan ^{-1} x=\frac{\pi}{4}-\tan ^{-1}\left(\frac{1}{3}\right)$
$$
\begin{aligned}
&\Rightarrow \tan ^{-1} x=\tan ^{-1}(1)-\tan ^{-1}\left(\frac{1}{3}\right) \\
&\left.\Rightarrow \quad \tan ^{-1} x=\tan ^{-1}\left(\frac{1-\frac{\pi}{3}}{1+\frac{1}{3}}\right) \tan ^{-1}(1)\right) \\
&\Rightarrow \quad\left[\because \tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)\right] \\
&\Rightarrow \quad \tan ^{-1} x=\tan ^{-1}\left(\frac{2 / 3}{4 / 3}\right) \\
&\Rightarrow \quad x=1 / 2
\end{aligned}
$$
Hence, the correct answer is option: 'B'.