<s>[INST] Solve inverse trigonometric equation $\frac{\pi}{6}=\tan^{-1} \frac{11}{x} -\tan^{-1} \frac{1}{x}$ How do I go about solving for $x$ when I have:
$\frac{\pi}{6}=\tan^{-1} \left( \frac{11}{x} \right)-\tan^{-1}\left( \frac{1}{x} \right)$. 
 [/INST] Apply the identity $\tan^{-1}a-\tan^{-1}b=\tan^{-1}\frac{a-b}{1+ab}$ to rewrite the equation 
$$\frac\pi6=\tan^{-1}\frac{11}{x} -\tan^{-1} \frac{1}{x}
=\tan^{-1}\frac{\frac{10}x}{1+\frac{11}{x^2}}$$
Then, take $\tan(\cdot)$ on both sides along with $\tan\frac\pi6=\frac1{\sqrt3}$,
$$\frac{11}{x^2}+\frac{10\sqrt3}x+1=0$$
which is a quadratic equation in $1/x$. Solve to obtain 
$$x=5\sqrt3\pm8$$
</s>