Q.Find the principal value of the following: $\cot^{-1} \left( \sqrt{3} \right)$

The key idea is that the principal value branch of $\cot^{-1}$ is $(0, \pi)$, and we need the angle $\theta$ in this interval whose cotangent equals $\sqrt{3}$. …

1. Understanding the inverse cotangent function

The function $\cot^{-1}(x)$ asks: "Which angle $\theta$ (in the principal value range) has cotangent equal to $x$?"

For $\cot^{-1}$, the standard principal value branch is $(0, \pi)$ — that is, all angles strictly between $0$ and $\pi$ radians, excluding $0$ and $\pi$ themselves (since $\cot$ is undefined at those points). This is different from $\tan^{-1}$, which uses $(-\pi/2, \pi/2)$. The reason for $(0, \pi)$ is that $\cot \theta$ is continuous and one-to-one on that interval, covering all real numbers exactly once.

So we need an angle $\theta$ such that:

• $\cot \theta = \sqrt{3}$
• $0 < \theta < \pi$

2. Relating $\cot$ to $\tan$

A common trick: $\cot \theta = \frac{1}{\tan \theta}$, provided $\tan \theta \neq 0$. So $\cot \theta = \sqrt{3}$ means $\tan \theta = \frac{1}{\sqrt{3}}$.

Now, $\tan \theta = \frac{1}{\sqrt{3}}$ is a familiar value. From standard angles:

• $\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}$

So $\theta = \frac{\pi}{6}$ is a candidate.

--- …

Method: Principal value of an inverse cotangent

Use this for $\cot^{-1}$ of a given number, reporting the unique angle in its branch.

Steps

Step 1: State the branch

$\cot^{-1}$ returns angles in the open interval $(0, \pi)$, where cotangent is strictly decreasing and covers every real value once.

Step 2: Relate cotangent to a known value

Either recall the standard cotangent directly or use $\cot\theta = \frac{1}{\tan\theta}$ to turn it into a tangent you know (e.g. $\cot\theta = \sqrt{3} \iff \tan\theta = \frac{1}{\sqrt{3}}$).

Step 3: Select the angle inside $(0,\pi)$ …

Common Mistakes

Mistake 1: Confusing which ratio gives $\sqrt{3}$ and answering $\frac{\pi}{3}$

Why it's wrong: $\cot\frac{\pi}{3} = \frac{1}{\sqrt{3}}$, while $\cot\frac{\pi}{6} = \sqrt{3}$. Correct approach: match the cotangent carefully — the angle with $\cot\theta = \sqrt{3}$ is $\frac{\pi}{6}$.

Mistake 2: Using the $\tan^{-1}$ branch $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ for $\cot^{-1}$ …