<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] Let $y = 1/x$.  Then solve 
$$\pi/6 = \tan^{-1} (11 y) - \tan^{-1} y$$
or
$$\pi/6 + \tan^{-1} y = \tan^{-1} (11 y)$$
to find $y =\frac{5 \sqrt{3}}{11}-\frac{8}{11}$, making $x$ easy to determine.
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