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

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

Use the identity:
$$\tan^{-1}a+\tan^{-1}b=\tan^{-1}\left(\frac{a+b}{1-ab}\right)$$).

So the equation becomes:
$$\tan^{-1}\left(\frac{10x}{1-24x^2}\right)=\frac{\pi}{6}$$

Taking tangent on both sides:
$$\frac{10x}{1-24x^2}=\frac{1}{\sqrt{3}}$$

Now solve:
$$10x=\frac{1-24x^2}{\sqrt{3}}$$
$$\Rightarrow10\sqrt{3},x=1-24x^2$$
$$\Rightarrow24x^2+10\sqrt{3}x-1=0$$

Use the quadratic formula:

$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$

Here (a=24),$$(b=10\sqrt{3})$$, (c=-1):
$$x=\frac{-10\sqrt{3}\pm\sqrt{(10\sqrt{3})^2+96}}{48}$$
=$$\frac{-10\sqrt{3}\pm6\sqrt{11}}{48}$$
= $$\frac{-5\sqrt{3}\pm3\sqrt{11}}{24}$$

Now check the interval

$$\quad-\frac{1}{2\sqrt{6}}$$ < x <$$\frac{1}{2\sqrt{6}}$$

• $$(x=\frac{-5\sqrt{3}+3\sqrt{11}}{24}\approx0.054)$$✅ inside
• $$(x=\frac{-5\sqrt{3}-3\sqrt{11}}{24}\approx-0.775)$$❌ outside
  

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