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

Concept: Inverse Trigonometric Graphs — principal value ranges matter, especially for $\sec^{-1}$ where the range is $[0,\pi]$, $\frac{\pi}{2}$ excluded, and negative inputs fall in $(\frac{\pi}{2}, \pi]$.

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 get

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

Step 2: Evaluate $\sec^{-1}(-2)$.

We need $\theta \in [0,\pi]$, $\theta \neq \frac{\pi}{2}$, such that $\sec\theta = -2$, i.e. $\cos\theta = -\frac12$. …

Let’s start with the core idea. Inverse trigonometric functions are not the same as their ordinary counterparts — they are defined only on specific intervals (principal value branches) so that they give a single, unique output. For $\tan^{-1}$, the principal value range is $(-\frac{\pi}{2}, \frac{\pi}{2})$, and for $\sec^{-1}$, it is $[0, \pi]$ excluding $\frac{\pi}{2}$. The trick is to find the angle in these ranges that matches the given input.

1. Evaluate $\tan^{-1} \sqrt{3}$ We need an angle $\theta$ in $(-\frac{\pi}{2}, \frac{\pi}{2})$ such that $\tan \theta = \sqrt{3}$. Since $\tan \frac{\pi}{3} = \sqrt{3}$ and $\frac{\pi}{3}$ lies inside $(-\frac{\pi}{2}, \frac{\pi}{2})$, we get:

2. Evaluate $\sec^{-1}(-2)$ This is trickier because the input is negative. The principal value branch for $\sec^{-1}$ is $[0, \pi]$, with $\sec^{-1}(x) = \cos^{-1}\left(\frac{1}{x}\right)$ for $|x| \geq 1$. So we need an angle $\phi$ in $[0, \pi]$ (excluding $\frac{\pi}{2}$) such that $\sec \phi = -2$, i.e., $\cos \phi = -\frac{1}{2}$. In $[0, \pi]$, $\cos \phi = -\frac{1}{2}$ occurs at $\phi = \frac{2\pi}{3}$ (since $\cos \frac{2\pi}{3} = -\frac{1}{2}$). Therefore: $\sec^{-1}(-2) = \frac{2\pi}{3}$ …

Method: Evaluating a difference of inverse trig values

Use this for questions that combine two inverse trig terms; evaluate each in its branch, then subtract.

Steps

Step 1: Evaluate the straightforward term

$\tan^{-1}$ of a positive standard value is a first-quadrant angle in $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$.

Step 2: Handle the inverse-secant term via cosine …

Common Mistakes

Mistake 1: Taking $\sec^{-1}(-2) = \frac{\pi}{3}$ because $\sec\frac{\pi}{3} = 2$

Why it's wrong: this ignores the minus sign; $\sec\theta = -2$ means $\cos\theta = -\frac{1}{2}$, giving a second-quadrant angle. Correct approach: $\sec^{-1}(-2) = \frac{2\pi}{3}$.

Mistake 2: Assuming $\sec^{-1}$ is odd and writing $-\frac{\pi}{3}$ for it …