Question

Let $f(x) = \frac{1}{1 + \tan x}, 0 < x < \frac{\pi}{2}$. Then $f^{-1}(x) =$}

• A. $\frac{1 - \tan x}{\tan x}$
• B. $\tan^{-1}(\frac{x}{1+x})$
• C. $\tan^{-1}(\frac{1-x}{x})$
• D. $\tan^{-1}(\frac{x}{1-x})$

Hint:

When finding inverses of fractional functions, use basic cross-multiplication. If the result involves a trigonometric term, simply isolate it and apply the inverse trigonometric operation at the end.

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
To find the inverse function \(f^{-1}(x)\), we set \(y = f(x)\), solve algebraically for \(x\) in terms of \(y\), and then swap \(x\) and \(y\).
Step 2: Key Formula or Approach:
Let \(y = \frac{1}{1 + \tan x}\).
Isolate \(\tan x\).
Apply \(\tan^{-1}\) to both sides to solve for \(x\).
Step 3: Detailed Explanation:
Let \(y = \frac{1}{1 + \tan x}\).
Multiply both sides by \((1 + \tan x)\):
\[ y(1 + \tan x) = 1 \] Divide by \(y\):
\[ 1 + \tan x = \frac{1}{y} \] Subtract 1 from both sides:
\[ \tan x = \frac{1}{y} - 1 \] Get a common denominator on the right side:
\[ \tan x = \frac{1 - y}{y} \] Take the inverse tangent of both sides to isolate \(x\):
\[ x = \tan^{-1}\left(\frac{1 - y}{y}\right) \] To write the inverse function in terms of \(x\), replace \(y\) with \(x\):
\[ f^{-1}(x) = \tan^{-1}\left(\frac{1 - x}{x}\right) \] Step 4: Final Answer:
The inverse function is \(\tan^{-1}\left(\frac{1 - x}{x}\right)\).