Discussion
Let \(f(x) = \dfrac{1}{1 + \tan x}\), \(0 \lt x \lt \frac{\pi}{2}\). Then \(f^{-1}(x) =\)
Answer: C
\(f^{-1}(x) = \tan^{-1}\left(\frac{1 - x}{x}\right)\).
Put \(y = \frac{1}{1 + \tan x}\) and solve for \(x\):
\[ \begin{aligned} 1 + \tan x &= \frac{1}{y} \\ \tan x &= \frac{1 - y}{y} \\ x &= \tan^{-1}\left(\frac{1 - y}{y}\right) \end{aligned} \]
Renaming \(y\) as \(x\) gives the inverse. For \(0 \lt x \lt \frac{\pi}{2}\), \(\tan x \gt 0\) so \(0 \lt y \lt 1\) and \(\tan^{-1}\) returns the right angle. The options \(\frac{1 - \tan x}{\tan x}\) and \(\frac{\tan x}{1 - \tan x}\) are not inverses at all, only rearrangements of \(\tan x\).
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