Question

Solve the following equation
$tan^{-1}\,2x + tan^{-1}\,3x = \frac{\pi}{4}$

• A. $1$
• B. $\frac{1}{6}$
• C. $\frac{1}{3}$
• D. $-\frac{1}{6}$

Answer 1

Answer: (B)

We have, $tan^{-1}\,2x + tan^{-1}\,3x = \frac{\pi}{4}$
$\Rightarrow tan^{-1}\left\{\frac{2x+3x}{1-2x \times 3x}\right\}= tan^{-1}\,1,$
if $6x^{2} < 1$
$\Rightarrow \frac{5x}{1-6x^{2}} = 1$,
if $6x^{2} < 1$
$\Rightarrow 6x + 5 x - 1 = 0$ and $x^{2} < \frac{1}{6}$
$\Rightarrow \left(6 x -1\right)\left(x + 1\right) = 0$ and $-\frac{1}{\sqrt{6}} < x < \frac{1}{\sqrt{6}}$
$\Rightarrow x = -1, \frac{1}{6}$ and $-\frac{1}{\sqrt{6}} < x < \frac{1}{\sqrt{6}}$
$\Rightarrow x = \frac{1}{6}$