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

Concept: Inverse Cotangent Principal Value — For $\cot^{-1}(x)$, the principal value range is $(0, \pi)$. We find the angle $\theta$ in $(0, \pi)$ such that $\cot \theta = -\frac{1}{\sqrt{3}}$.

Step 1: Recall that $\cot \theta = \frac{1}{\tan \theta}$, so $\cot \theta = -\frac{1}{\sqrt{3}}$ means $\tan \theta = -\sqrt{3}$. …

Why the range matters

When we talk about "principal value" of an inverse trigonometric function, we mean the unique angle in a specific interval that gives the required trigonometric ratio. For $\cot^{-1}$, the standard principal value range is $(0, \pi)$ — that is, angles strictly between $0$ and $\pi$ radians.

Why this range? Because cotangent is one-to-one on $(0, \pi)$, so every real number appears exactly once as a cotangent value in that interval. This lets us define a proper inverse function.

The key point: the principal value must lie in $(0, \pi)$, not in $(-\pi/2, \pi/2)$ like for $\tan^{-1}$. This is a common source of mistakes.

Step-by-step

1. Identify the problem. We need an angle $\theta$ such that $\cot \theta = -\frac{1}{\sqrt{3}}$ and $0 < \theta < \pi$.

2. Recall the cotangent of standard angles.

   $\cot \frac{\pi}{3} = \frac{1}{\sqrt{3}}$, and $\cot \frac{2\pi}{3} = -\frac{1}{\sqrt{3}}$.

   Both $\frac{\pi}{3}$ and $\frac{2\pi}{3}$ lie in $(0, \pi)$, but only one gives the negative value we need.

3. Check the sign.

   Since $\cot \theta = \frac{\cos\theta}{\sin\theta}$, the sign of $\cot \theta$ depends on the quadrant:

   • In $(0, \frac{\pi}{2})$ (Quadrant I): $\cos\theta > 0$, $\sin\theta > 0$ → $\cot\theta > 0$.
   • In $(\frac{\pi}{2}, \pi)$ (Quadrant II): $\cos\theta < 0$, $\sin\theta > 0$ → $\cot\theta < 0$.

   Our target value is negative, so $\theta$ must be in Quadrant II.

4. Find the specific angle.

   The reference angle for $\cot^{-1}\left(\frac{1}{\sqrt{3}}\right)$ is $\frac{\pi}{3}$.

   In Quadrant II, the angle with that reference is $\pi - \frac{\pi}{3} = \frac{2\pi}{3}$. …

Method: Principal value of an inverse function with a negative argument

Use this for $\cot^{-1}$, $\cos^{-1}$ or any inverse whose principal range is $[0,\pi]$ or $(0,\pi)$ (not symmetric about $0$), especially when the argument is negative.

Steps

Step 1: State the correct principal-value range for this specific function.

Do not reuse $\tan^{-1}$'s range by habit. For inverse cotangent,

Because this range never dips below $0$, a negative-argument answer must be an obtuse angle, not a negative one.

Step 2: Find the reference (acute) angle from the magnitude. …

Common Mistakes

Mistake 1: Using $\tan^{-1}$'s range and answering $-\dfrac{\pi}{3}$.

Why it's wrong: students treat $\cot^{-1}$ like $\tan^{-1}$ and return a negative acute angle, but the principal range of $\cot^{-1}$ is $(0,\pi)$, which contains no negative angle. Correct approach: for a negative argument the answer must be obtuse — take $\pi-\tfrac{\pi}{3}=\tfrac{2\pi}{3}$.

Mistake 2: Assuming $\cot^{-1}(-y)=-\cot^{-1}(y)$. …