Step 1: Understanding the Concept:
Trigonometric functions are periodic, meaning they repeat their values over regular intervals. Because of this, inverse trigonometric functions like \(\tan^{-1}(x)\) could potentially have infinitely many values for a single input.
To make them true functions, we restrict their range to a specific interval called the Principal Branch.
The principal value is the specific angle that lies within this restricted range.
For the function \(y = \tan^{-1}(x)\), the principal value branch is:
\[ -\frac{\pi}{2}<y<\frac{\pi}{2} \quad \text{or} \quad y \in (-\frac{\pi}{2}, \frac{\pi}{2}) \]
Step 2: Detailed Explanation:
Let the required principal value be \(\theta\).
Therefore, we have:
\[ \theta = \tan^{-1}(1) \]
This can be rewritten as:
\[ \tan(\theta) = 1 \]
We are looking for an angle \(\theta\) such that its tangent is \(1\), and it must fall within the interval \((-\frac{\pi}{2}, \frac{\pi}{2})\).
Recall the tangent values for standard angles in the first quadrant:
\(\tan(0) = 0\)
\(\tan(30^\circ) = \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}}\)
\(\tan(45^\circ) = \tan(\frac{\pi}{4}) = 1\)
\(\tan(60^\circ) = \tan(\frac{\pi}{3}) = \sqrt{3}\)
Since \(\tan(\frac{\pi}{4}) = 1\) and the value \(\frac{\pi}{4}\) (which is \(45^\circ\)) is clearly between \(-\frac{\pi}{2}\) (\(-90^\circ\)) and \(\frac{\pi}{2}\) (\(90^\circ\)), it satisfies all conditions for being the principal value.
Step 3: Final Answer:
The principal value of \(\tan^{-1}(1)\) is \(\frac{\pi}{4}\).