Question

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

• A. $\frac{1}{6}$
• B. $-1$
• C. $\left(\frac{1}{6},\,-1\right)$
• D. None of these

Answer 1

Answer: (A)

$tan^{-1}\,2x+tan^{-1}\,3x = \frac{\pi}{4}$
$\Rightarrow \frac{3x+2x}{1 - 6x^{2}} = tan \frac{\pi }{4}$
$\Rightarrow 5x = 1 - 6x^{2}$
$\Rightarrow 6x^{2} + 5 x - 1 = 0$
$\Rightarrow x = -1, \frac{1}{6}$
But when $x = -1$, $tan^{-1}2x = tan^{-1} \left(-2\right) < 0$ and $tan^{-1}3x = tan^{-1}\left(-3\right) < 0$
This value will not satisfy the given equation.
Hence, $x = \frac{1}{6}$.