Question

If ${\tan }^{-1}2x+{\tan }^{-1}3x=\frac{\pi }{4}$, then $x$ is equal to

• A. $-1$
• B. $-1,\frac{1}{10}$
• C. $1,\frac{1}{6}$
• D. None of these

Answer 1

Answer: (C)

Given, ${\tan }^{-1}2x+{\tan }^{-1}3x=\frac{\pi }{4}.....$(i)
$\Rightarrow {\tan }^{-1}\left(\frac{2x+3x}{1-2x\cdot 3x}\right)=\frac{\pi }{4}$
$\Rightarrow {\tan }^{-1}\left(\frac{5x}{1-6x^2}\right)=\frac{\pi }{4}$
$\Rightarrow \frac{5x}{1-6x^2}=\tan \frac{\pi }{4}$
$\Rightarrow \frac{5x}{1-6x^2}=1$
$\Rightarrow 5x=1-6x^2$
$\Rightarrow 6x^2+5x-1=0$
$\Rightarrow 6x^2+6x-x-1=0$
$\Rightarrow 6x(x+1)-1(x+1)=0$
$\Rightarrow (x+1)(6x-1)=0\Rightarrow x=\frac{1}{6},-1$
But $x=-1$ does not satisfy Eq. (i). Since here RHS $>0$ and ${\tan }^{-1}(x)<0$ for $x<0$.
So, $x=\frac{1}{6}$ is the only solution for given equation.