Mark the tick against the correct answer in the following:
If $\tan ^{-1} 3 \mathrm{x}+\tan ^{-1} 2 \mathrm{x}=\frac{\pi}{4}$ then $\mathrm{x}=?$
A. $\frac{1}{2}$ or $-2$
B. $\frac{1}{3}$ or $-3$
C. $\frac{1}{4}$ or $-2$
D. $\frac{1}{6}$ or $-1$
Given: $\tan ^{-1} 3 x+\tan ^{-1} 2 x=\frac{\pi}{4}$
To Find: The value of $x$
Since we know that $\tan ^{-1} x+\tan ^{-1} y=\tan ^{-1}\left(\frac{x+y}{1-x y}\right)$
$\Rightarrow \tan ^{-1} 3 x+\tan ^{-1} 2 x=\tan ^{-1}\left(\frac{3 x+2 x}{1-(3 x \times 2 x)}\right)$
$=\tan ^{-1}\left(\frac{5 x}{1-6 x^{2}}\right)$
Now since $\tan ^{-1} 3 x+\tan ^{-1} 2 x=\frac{\pi}{4}$
$\tan ^{-1} 3 x+\tan ^{-1} 2 x=\tan ^{-1} 1\left(\cdots \tan \frac{\pi}{4}=1\right)$
$\Rightarrow \tan ^{-1}\left(\frac{5 x}{1-6 x^{2}}\right)=\tan ^{-1} 1$
$\Rightarrow \frac{5 x}{1-6 x^{2}}=1$
$\Rightarrow 6 x^{2}+5 x-1=0$
$\Rightarrow x=\frac{1}{6}$ or $x=-1$
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