Mark the tick against the correct answer in the following:
If $\tan ^{-1} \mathrm{x}=\frac{\pi}{4}-\tan ^{-1} \frac{1}{3}$ then $\mathrm{x}=?$
A. $\frac{1}{2}$
B. $\frac{1}{4}$
C. $\frac{1}{6}$
D. None of these
To Find: The value of $\tan ^{-1} x=\frac{\pi}{4}-\tan ^{-1} \frac{1}{3}$
Now, $\tan ^{-1} x=\tan ^{-1} 1-\tan ^{-1} \frac{1}{3}\left(\because \tan \frac{\pi}{4}=1\right)$
Since we know that $\tan ^{-1} x-\tan ^{-1} y=\tan ^{-1}\left(\frac{x-y}{1+x y}\right)$
$\Rightarrow \tan ^{-1} 1+\tan ^{-1} \frac{1}{3}=\tan ^{-1}\left(\frac{1-\frac{1}{3}}{1+\frac{1}{3}}\right)=\tan ^{-1} \frac{1}{2}$
$\Rightarrow \tan ^{-1} x=\tan ^{-1} \frac{1}{2}$
$\Rightarrow x=\frac{1}{2}$
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