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:
$\frac{\pi}{6}=\tan^{-1} \left( \frac{11}{x} \right)-\tan^{-1}\left( \frac{1}{x} \right)$. 
 A: $30^\circ = \tan^{-1}\frac {11}x - \tan^{-1} \frac {1}{x}$
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:
$x = 5\sqrt 3 \pm \sqrt {75 - 11}\\
x = 5\sqrt 3 \pm 8$ 
A: 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 
$$x=5\sqrt3\pm8$$
A: 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.
A: 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.
A: The formula for the tangent of a difference says
$$
\tan(a-b) = \frac{\tan a-\tan b}{1+\tan a \tan b}.
$$
If $p = \tan a$ and $q=\tan b$ and $a = \tan^{-1}p$ and $b=\tan^{-1} q$ then we have
$$
\tan(a-b) = \frac{p-q}{1+pq}
$$
so
$$
a-b = \tan^{-1} \frac{p-q}{1+pq}
$$
or in other words
$$
\tan^{-1} p - \tan^{-1} q = \tan^{-1}\frac{p-q}{1+pq}.
$$
So plug in the numbers you specified:
$$
\tan^{-1} \frac {11} x - \tan^{-1} \frac 1 x = \tan^{-1} \frac{\frac{11}x - \frac 1 x}{1 + \left(\frac{11} x \cdot \frac 1 x\right)}
$$
$$
= \tan^{-1} \frac{10x}{x^2+11}
$$
Thus
$$
30^\circ = \tan^{-1} \frac{10x}{x^2+11}
$$
$$
\frac{10x}{x^2+11} = \tan 30^\circ = \frac{\sqrt 3}3
$$
$$
30x = \sqrt 3\, x^2 + 11\sqrt 3.
$$
This is a quadratic equation.
