Q.The principal value of $\tan^{-1}\sqrt3$ is __________.

The key idea is that the principal value of $\tan^{-1}x$ lies in the open interval $(-\frac{\pi}{2}, \frac{\pi}{2})$.

We need the angle $\theta$ such that $\tan\theta = \sqrt{3}$ and $\theta$ is in $(-\frac{\pi}{2}, \frac{\pi}{2})$. …

Why the principal value matters

When you see $\tan^{-1} x$, it asks: which angle $\theta$ has tangent equal to $x$? But here’s the catch — the tangent function is periodic, repeating every $\pi$ radians. So there are infinitely many angles with the same tangent. For example, $\tan\frac{\pi}{3} = \sqrt{3}$, but also $\tan\frac{4\pi}{3} = \sqrt{3}$, $\tan\frac{7\pi}{3} = \sqrt{3}$, and so on.

To make $\tan^{-1}$ a proper function (one input gives exactly one output), we restrict the range to a specific interval called the principal value branch. For $\tan^{-1}$, that branch is $(-\frac{\pi}{2}, \frac{\pi}{2})$ — the open interval between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$.

Step-by-step

1. Identify the equation. We want $\theta = \tan^{-1}\sqrt{3}$. By definition, this means $\tan\theta = \sqrt{3}$ and $\theta$ must lie in the principal value branch $(-\frac{\pi}{2}, \frac{\pi}{2})$. …

Method: Evaluating $\tan^{-1}$ of a standard value

Steps

Step 1: Recall the range $\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$.

Step 2: Match the value to a standard angle.

Find $\theta$ whose tangent equals the given number from the standard table: $\tan\tfrac{\pi}{6}=\tfrac{1}{\sqrt3}$, $\tan\tfrac{\pi}{4}=1$, $\tan\tfrac{\pi}{3}=\sqrt3$. …

Common Mistakes

Mistake 1: Confusing $\sqrt3$ with $\frac{1}{\sqrt3}$.

Why it's wrong: $\tan\frac{\pi}{6}=\frac{1}{\sqrt3}$, not $\sqrt3$. Correct approach: $\tan\frac{\pi}{3}=\sqrt3$, so $\tan^{-1}\sqrt3=\frac{\pi}{3}$.

Mistake 2: Offering $\frac{4\pi}{3}$. …