Inverse Tangent Identity
There are real numbers $ a $ and $ b $ such that for every positive number $ x $, we have the identity
\[
\tan^{-1} \left( \frac{1}{x} - \frac{x}{8} \right) + \tan^{-1}(ax) + \tan^{-1}(bx)
= \frac{\pi}{2} \, .\]
(Throughout this equation, $ \tan^{-1} $ means the inverse tangent function, sometimes written $ \arctan $.) What is the value of $ a^2 + b^2 $?
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- $\frac{a}{b}$
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Solution
By the tangent addition formula,
\[\tan (\tan^{-1} p + \tan^{-1} q) = \frac{p + q}{1 - pq}.\]From the given equation,
\[\tan^{-1} ax + \tan^{-1} bx = \frac{\pi}{2} - \tan^{-1} \left( \frac{1}{x} - \frac{x}{8} \right).\]Then
\[\tan (\tan^{-1} ax + \tan^{-1} bx) = \tan \left( \frac{\pi}{2} - \tan^{-1} \left( \frac{1}{x} - \frac{x}{8} \right) \right),\]The left-hand side is $ \frac{ax + bx}{1 - abx^2} $. The right-hand side is
\begin{align*}
\tan \left( \frac{\pi}{2} - \tan^{-1} \left( \frac{1}{x} - \frac{x}{8} \right) \right) &= \frac{1}{\tan \left( \tan^{-1} \left( \frac{1}{x} - \frac{x}{8} \right) \right)} \\
&= \frac{1}{\frac{1}{x} - \frac{x}{8}} \\
&= \frac{x}{1 - \frac{1}{8} x^2}.\end{align*}Hence, $ a + b = 1 $ and $ ab = \frac{1}{8}, $ so
\[a^2 + b^2 = (a + b)^2 - 2ab = 1 - \frac{2}{8} = \boxed{\frac{3}{4}}.\]
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