Question

$\tan^{- 1}4x + \tan^{- 1}6x = \frac{\pi}{6}$

$${\Rightarrow \tan^{- 1}\left( \frac{4x + 6x}{1 + 24x^{2}} \right) = \frac{\pi}{6} }{\Rightarrow \frac{10x}{1 - 24x^{2}} = \frac{1}{\sqrt{3}} }{\Rightarrow 24x^{2} + 10\sqrt{3}x - 1 = 0 }{x = \frac{- 10\sqrt{3} \pm \sqrt{300 + 96}}{48} }{x = \frac{\sqrt{396} - 10\sqrt{3}}{48} }$$Only 1 solution in $\left( - \frac{1}{2\sqrt{6}},\frac{1}{2\sqrt{6}} \right)$