<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] $30^\circ = \tan^{-1}\frac {11}x - \tan^{-1} \frac {1}{x}$
Take the tan of both sides
$\tan 30^\circ = \tan (\tan^{-1}\frac {11}x - \tan^{-1} \frac {1}{x})$
Angle addition - subtraction rule for tangent
$\tan(A+B) = \frac {\tan A+ \tan B}{1-\tan A\tan B}\\
\tan(A-B) = \frac {\tan A- \tan B}{1+\tan A\tan B}$
$\frac 1{\sqrt 3} = \frac {\tan (\tan^{-1} \frac {11}{x}) - \tan (\tan ^{-1}\frac {1}{x})}{1 + \tan (\tan^{-1} \frac {11}{x})\tan(\tan^{-1} \frac {1}{x})} $
$\tan (\tan^{-1} y) = y$
$\frac 1{\sqrt 3} = \frac {\frac {11}{x} - \frac{1}{x}}{1 + \frac {11}{x^2}} $
Multiply numerator and denominator by $x^2$ to kill the fractions.
$\frac 1{\sqrt 3} = \frac {10x}{x^2 + 11} $
$x^2 + 11 = 10\sqrt 3 x\\
x^2 - 10\sqrt3 x + 11 = 0$
Use the quadratic formula:
$x = 5\sqrt 3 \pm \sqrt {75 - 11}\\
x = 5\sqrt 3 \pm 8$ 
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