Question

The number of value(s) of $x$ satisfying the equation ${\tan }^{-1}2x+{\tan }^{-1}3x=\frac{\pi }{4}$ is

• A. infinitely many
• B. $0$
• C. $1$
• D. $2$

Answer 1

The correct option is C $1$

We know,
${\tan }^{-1}a+{\tan }^{-1}b=\{\begin{array}{cc}{\tan }^{-1}\frac{a+b}{1-ab}& \text{if}a,b>0,ab<1\\ \pi +{\tan }^{-1}\frac{a+b}{1-ab}& \text{if}a,b>0,ab>1\end{array}$

In the problem, second case is not possible as the minimum value of $\pi +{\tan }^{-1}\frac{a+b}{1-ab}$ can be $\frac{\pi }{2}.$
So it can not be equal to $\frac{\pi }{4}$

Now,
$6x^2<1\Rightarrow x\in (-\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}})$
Solving the equation
${\tan }^{-1}2x+{\tan }^{-1}3x=\frac{\pi }{4}\Rightarrow {\tan }^{-1}\left(\frac{2x+3x}{1-6x^2}\right)=\frac{\pi }{4}\Rightarrow \frac{5x}{1-6x^2}=1$
$\Rightarrow 6x^2+5x-1=0\Rightarrow (6x-1)(x+1)=0\Rightarrow x=\frac{1}{6}(\because x\in (-\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}}))$
$\therefore$ There is only one solution.