Tangent Equation Solutions
Find the solutions to
\[\frac{1}{x - \tan 20^{\circ}} + \frac{1}{x + \tan 40^{\circ}} + \frac{1}{x - \tan 80^{\circ}} = 0.\]Enter the solutions, separated by commas.
- 1
- 2
- 3
- +
- 4
- 5
- 6
- -
- 7
- 8
- 9
- $\frac{a}{b}$
- .
- 0
- =
- %
- $a^n$
- $a^{\circ}$
- $a_n$
- $\sqrt{}$
- $\sqrt[n]{}$
- $\pi$
- $\ln{}$
- $\log$
- $\theta$
- $\sin{}$
- $\cos{}$
- $\tan{}$
- $($
- $)$
- $[$
- $]$
- $\cap$
- $\cup$
- $,$
- $\infty$
Solution
Let $ a = \tan 20^\circ, $ $ b = \tan 40^\circ, $ and $ c = \tan 80^\circ, $ so
\[\frac{1}{x - a} + \frac{1}{x + b} + \frac{1}{x - c} = 0.\]Then $ (x + b)(x - c) + (x - a)(x - c) + (x - a)(x + b) = 0, $ which expands as
\[3x^2 + (-2a + 2b - 2c) x + (-ab + ac - bc) = 0.\]Let $ t = \tan 10^\circ $. Then from the addition formula for tangent,
\begin{align*}
-a + b - c &= -\tan 20^\circ + \tan 40^\circ - \tan 80^\circ \\
&= -\tan (30^\circ - 10^\circ) + \tan (30^\circ + \tan 10^\circ) - \frac{1}{\tan 10^\circ} \\
&= -\frac{\tan 30^\circ - \tan 10^\circ}{1 + \tan 30^\circ \tan 10^\circ} + \frac{\tan 30^\circ + \tan 10^\circ}{1 - \tan 30^\circ \tan 10^\circ} - \frac{1}{\tan 10^\circ} \\
&= -\frac{\frac{1}{\sqrt{3}} - t}{1 + \frac{t}{\sqrt{3}}} + \frac{\frac{1}{\sqrt{3}} + t}{1 - \frac{t}{\sqrt{3}}} - \frac{1}{t} \\
&= -\frac{1 - t \sqrt{3}}{\sqrt{3} + t} + \frac{1 + t \sqrt{3}}{\sqrt{3} - t} - \frac{1}{t} \\
&= -\frac{(1 - t \sqrt{3})(\sqrt{3} - t)}{3 - t^2} + \frac{(1 + t \sqrt{3})(\sqrt{3} + t)}{3 - t^2} - \frac{1}{t} \\
&= \frac{8t}{3 - t^2} - \frac{1}{t} \\
&= \frac{9t^2 - 3}{3t - t^3}.\end{align*}By the triple angle formula,
\[\frac{1}{\sqrt{3}} = \tan 30^\circ = \tan (3 \cdot 10^\circ) = \frac{3t - t^3}{1 - 3t^2},\]so $ \frac{1 - 3t^2}{3t - t^3} = \sqrt{3} $. Then
\[\frac{9t^2 - 3}{3t - t^3} = -3 \sqrt{3},\]so $ -2a + 2b - 2c = -6 \sqrt{3} $.
Also,
\begin{align*}
-ab + ac - bc &= -\tan 20^\circ \tan 40^\circ + \tan 20^\circ \tan 80^\circ - \tan 40^\circ \tan 80^\circ \\
&= -\frac{1 - t \sqrt{3}}{\sqrt{3} + t} \cdot \frac{1 + t \sqrt{3}}{\sqrt{3} - t} + \frac{1 - t \sqrt{3}}{\sqrt{3} + t} \cdot \frac{1}{t} - \frac{1 + t \sqrt{3}}{\sqrt{3} - t} \cdot \frac{1}{t} \\
&= -\frac{1 - 3t^2}{3 - t^2} + \frac{1}{t} \left( \frac{1 - t \sqrt{3}}{\sqrt{3} + t} - \frac{1 + t \sqrt{3}}{\sqrt{3} - t} \right) \\
&= -\frac{1 - 3t^2}{3 - t^2} + \frac{1}{t} \cdot \left( -\frac{8t}{3 - t^2} \right) \\
&= \frac{3t^2 - 1}{3 - t^2} - \frac{8}{3 - t^2} \\
&= \frac{3t^2 - 9}{3 - t^2} \\
&= -3.\end{align*}Thus, the quadratic is
\[3x^2 - 6 \sqrt{3} x - 3 = 0.\]By the quadratic formula, the roots are $ \boxed{2 + \sqrt{3}, -2 + \sqrt{3}} $.
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