The number of solutions of tan^{-1}4x + tan^{ | vedclass

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(C) Given equation: $	an^{-1}4x + 	an^{-1}6x = \frac{\pi}{6}$.Using the formula $	an^{-1}A + 	an^{-1}B = 	an^{-1}\left(\frac{A+B}{1-AB}ight)$,

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(C) Given equation: $	an^{-1}4x + 	an^{-1}6x = \frac{\pi}{6}$.Using the formula $	an^{-1}A + 	an^{-1}B = 	an^{-1}\left(\frac{A+B}{1-AB}ight)$, we have:$	an^{-1}\left(\frac{4x+6x}{1-(4x)(6x)}ight) = \frac{\pi}{6}$.$\frac{10x}{1-24x^2} = 	an\left(\frac{\pi}{6}ight) = \frac{1}{\sqrt{3}}$.$10\sqrt{3}x = 1 - 24x^2 \implies 24x^2 + 10\sqrt{3}x - 1 = 0$.Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$:$x = \frac{-10\sqrt{3} \pm \sqrt{(10\sqrt{3})^2 - 4(24)(-1)}}{2(24)} = \frac{-10\sqrt{3} \pm \sqrt{300 + 96}}{48} = \frac{-10\sqrt{3} \pm \sqrt{396}}{48} = \frac{-10\sqrt{3} \pm 6\sqrt{11}}{48} = \frac{-5\sqrt{3} \pm 3\sqrt{11}}{24}$.The range is $-\frac{1}{2\sqrt{6}} $x_1 = \frac{-5\sqrt{3} + 3\sqrt{11}}{24} \approx \frac{-8.66 + 9.95}{24} \approx 0.054$ (In range).$x_2 = \frac{-5\sqrt{3} - 3\sqrt{11}}{24} \approx \frac{-8.66 - 9.95}{24} \approx -0.775$ (Not in range).Only one solution satisfies the condition.

The number of solutions of $\tan^{-1}4x + \tan^{-1}6x = \frac{\pi}{6}$ where $-\frac{1}{2\sqrt{6}} < x < \frac{1}{2\sqrt{6}}$ is equal to

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