Question

Let $f : \mathbb{R} \to \mathbb{R}$ be given by $f(x) = \tan x$. Then $f^{-1}(1)$ is:

• A. $\frac{\pi}{4}$
• B. $n\pi + \frac{\pi}{4}; \, n \in \mathbb{Z}$
• C. $\frac{\pi}{3}$
• D. $n\pi + \frac{\pi}{3}; \, n \in \mathbb{Z}$

Answer 1

The Correct Option is A

Step 1: Find the inverse of \( f(x) = \tan x \)

The function \( f(x) = \tan x \) is periodic, with a period of \( \pi \). The inverse function \( f^{-1}(x) \) gives us the angle whose tangent is \( x \).

Step 2: Find \( f^{-1}(1) \)

We need to find \( x \) such that:

\[ \tan x = 1. \]

The solution to \( \tan x = 1 \) is \( x = \frac{\pi}{4} + n\pi \), where \( n \in \mathbb{Z} \) because the tangent function repeats every \( \pi \).

Step 3: Conclusion

Thus, \( f^{-1}(1) = \{ n\pi + \frac{\pi}{4} : n \in \mathbb{Z} \} \).

Answer 2

The function \( f(x) = \tan x \) is a periodic function with a period of \( \pi \). 

The range of the tangent function is all real numbers, i.e., \( (-\infty, \infty) \). 

However, the function is not one-to-one (injective) on its entire domain, so it does not have an inverse function on the whole real line.

To find the inverse function, we restrict the domain of the tangent function to the interval \( \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \). On this interval, the tangent function is strictly increasing and has a range of \( (-\infty, \infty) \). 

The inverse function is denoted as \( \arctan(x) \) or \( \tan^{-1}(x) \).

The principal value of \( x \) for which \( \tan x = 1 \) is \( x = \frac{\pi}{4} \).

However, since the tangent function is periodic with period \( \pi \), there are infinitely many values of \( x \) for which \( \tan x = 1 \).

These values are given by \( x = n\pi + \frac{\pi}{4} \), where \( n \) is an integer.

Therefore, \( f^{-1}(1) \) is the set \( \{n\pi + \frac{\pi}{4} : n \in \mathbb{Z}\} \).