Question

Given \( y = \frac{1}{1 + \tan x} \), find \( f^{-1}(x) \), where \( 0<x<\frac{\pi}{2} \).

Hint:

To find the inverse of a function involving trigonometric functions, solve the equation for \( x \), then apply the inverse function.

Answer 1

Step 1: Express \( y \) in terms of \( x \).
We are given the equation: \[ y = \frac{1}{1 + \tan x} \] To find the inverse, we first solve for \( x \) in terms of \( y \).
Step 2: Rearrange the equation to isolate \( \tan x \).
Multiply both sides by \( 1 + \tan x \): \[ y(1 + \tan x) = 1 \] \[ y + y \tan x = 1 \] Now, subtract \( y \) from both sides: \[ y \tan x = 1 - y \] Finally, divide by \( y \) to solve for \( \tan x \): \[ \tan x = \frac{1 - y}{y} \]
Step 3: Take the inverse tangent to find \( x \).
Now, apply the inverse tangent to both sides: \[ x = \tan^{-1} \left( \frac{1 - y}{y} \right) \]
Step 4: Conclusion.
Thus, the inverse function is: \[ f^{-1}(x) = \tan^{-1} \left( \frac{1 - x}{x} \right) \]