<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] Make a right triangle ABC with C=90, AC=x, BC=11. Try to represent $\sin B$ in two ways: the straightforward way: 
$$
\sin B=\frac{x}{\sqrt{121+x^2}},
$$
and - let $D$ be on $BC$ with $CD=1$, apply  law of sine to triangle ADB and get 
$$ 
\frac{\sqrt{1+x^2}}{\sin B}=\frac{10}{\sin 30}=20,
$$ 
thus get 
$$
\frac{x}{\sqrt{121+x^2}}=\frac{\sqrt{1+x^2}}{20},
$$
square this we get $400x^2=(121+x^2)((1+x^2)$. You can let $y=x^2$ and that is a 2nd order equation you can solve.
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