Q.$\tan^{-1}\sqrt{3}-\cot^{-1}(-\sqrt{3})$ is equal to (A) $\pi$ (B) $-\frac{\pi}{2}$ (C) $0$ (D) $2\sqrt{3}$ Miscellaneous Examples

Concept: Inverse Tangent Identity — Use the principal value ranges of $\tan^{-1}$ and $\cot^{-1}$, and the relation $\cot^{-1}(-x) = \pi - \cot^{-1}(x)$ for real $x$.

Step 1: Evaluate $\tan^{-1}\sqrt{3}$.

Since $\tan\frac{\pi}{3} = \sqrt{3}$ and $\frac{\pi}{3}$ lies in the principal range $(-\frac{\pi}{2}, \frac{\pi}{2})$, we have

$\tan^{-1}\sqrt{3} = \frac{\pi}{3}$.

Step 2: Evaluate $\cot^{-1}(-\sqrt{3})$.

The principal range of $\cot^{-1}$ is $(0, \pi)$.

We know $\cot\frac{\pi}{6} = \sqrt{3}$, so $\cot^{-1}(\sqrt{3}) = \frac{\pi}{6}$.

Using the identity $\cot^{-1}(-x) = \pi - \cot^{-1}(x)$ for $x > 0$, …

The problem asks for $\tan^{-1}\sqrt{3} - \cot^{-1}(-\sqrt{3})$. At first glance, you might be tempted to compute each term directly — but the negative sign inside the $\cot^{-1}$ is the trap. The inverse cotangent function, like all inverse trigonometric functions, has a defined principal value range, and handling a negative argument requires care.

The core idea: use the identity that relates $\cot^{-1}(-x)$ to $\cot^{-1}x$. This lets you turn the negative argument into a positive one, after which you can evaluate both terms using standard angles.

1. Evaluate $\tan^{-1}\sqrt{3}$ We know $\tan\frac{\pi}{3} = \sqrt{3}$, and since $\frac{\pi}{3}$ lies in the principal range of $\tan^{-1}$ (which is $(-\frac{\pi}{2}, \frac{\pi}{2})$), we have:

2. Handle $\cot^{-1}(-\sqrt{3})$ using the negative-argument identity The principal value range for $\cot^{-1}$ is $(0, \pi)$. For any $x > 0$, the identity is:

This works because if $\cot^{-1}x = \theta$ where $0 < \theta < \pi$, then $\cot(\pi - \theta) = -\cot\theta = -x$, and $\pi - \theta$ also lies in $(0, \pi)$.

So with $x = \sqrt{3}$:

3. Find $\cot^{-1}\sqrt{3}$ Since $\cot\frac{\pi}{6} = \sqrt{3}$ and $\frac{\pi}{6}$ is in $(0, \pi)$, we get:

Therefore:

4. Subtract the two terms …

Method: Combining inverse-trig values with negative arguments

Use this for expressions like $\tan^{-1}(\cdot) \pm \cot^{-1}(\cdot)$ where one or more arguments is negative — the negative sign is the trap.

Steps

Step 1: Know each function's principal range.

$\tan^{-1}$ lands in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$; $\cot^{-1}$ lands in $(0, \pi)$. The ranges differ, so negative arguments behave differently.

Step 2: Convert negative arguments with the correct identity.

$\tan^{-1}(-x) = -\tan^{-1}x, \qquad \cot^{-1}(-x) = \pi - \cot^{-1}x.$ …

Common Mistakes

Mistake 1: Writing $\cot^{-1}(-\sqrt{3}) = -\frac{\pi}{6}$.

Why it's wrong: the principal range of $\cot^{-1}$ is $(0,\pi)$, so its output can never be negative. Correct approach: use $\cot^{-1}(-x) = \pi - \cot^{-1}x$, giving $\pi - \frac{\pi}{6} = \frac{5\pi}{6}$.

Mistake 2: Assuming $\cot^{-1}$ shares the range of $\tan^{-1}$. …