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How do I go about solving for $x$ when I have:

$\frac{\pi}{6}=\tan^{-1} \left( \frac{11}{x} \right)-\tan^{-1}\left( \frac{1}{x} \right)$.

Soham Konar
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I Lopez
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    This equation has no solutions. The range of arctangent is $(-\pi/2, \pi/2)$, so the right-hand side is in the range $(-\pi, \pi)$. There is no way the right-hand side can be as large as $30$. – Eric Towers Jan 18 '20 at 00:11
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    he means 30 degree – Yuval Jan 18 '20 at 00:22
  • Use $$-\tan^{-1}1/x=\tan^{-1}(-1/x)$$ and https://math.stackexchange.com/questions/1837410/inverse-trigonometric-function-identity-doubt-tan-1x-tan-1y-pi-tan – lab bhattacharjee Jan 18 '20 at 02:23

5 Answers5

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$30^\circ = \tan^{-1}\frac {11}x - \tan^{-1} \frac {1}{x}$

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:

$x = 5\sqrt 3 \pm \sqrt {75 - 11}\\ x = 5\sqrt 3 \pm 8$

user317176
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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

$$x=5\sqrt3\pm8$$

Quanto
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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.

David G. Stork
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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.

Yuval
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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.

Michael Hardy
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