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

$x = \frac{1}{6}$

The correct answer is Option (3) → $x = \frac{1}{6}$ ##

We have $\tan^{-1} 2x + \tan^{-1} 3x = \frac{\pi}{4}$

or $\quad \tan^{-1} \left( \frac{2x + 3x}{1 - 2x \times 3x} \right) = \frac{\pi}{4}$

i.e. $\quad \tan^{-1} \left( \frac{5x}{1 - 6x^2} \right) = \frac{\pi}{4}$

Therefore $\quad \frac{5x}{1 - 6x^2} = \tan \frac{\pi}{4} = 1$

or $\quad 6x^2 + 5x - 1 = 0 \text{ i.e., } (6x - 1)(x + 1) = 0$

which gives $\quad x = \frac{1}{6} \text{ or } x = -1$

Since $x = -1$ does not satisfy the equation, as the L.H.S. of the equation becomes negative, $x = \frac{1}{6}$ is the only solution of the given equation.