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

Question

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

Accepted Answer

> [!TLDR]
> The key is to evaluate each inverse trigonometric function within its principal value branch, then subtract carefully. The expression simplifies to $-\frac{\pi}{3}$, which corresponds to option (B).

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

2. **Evaluate $\sec^{-1}(-2)$**  
   This is trickier because the input is negative. The principal value branch for $\sec^{-1}$ is $[0, \pi]$, with $\sec^{-1}(x) = \cos^{-1}\left(\frac{1}{x}ight)$ 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:  
   $$ \sec^{-1}(-2) = \frac{2\pi}{3} $$ …

Inverse function	Domain (allowed inputs $x$ )	Principal value branch (angles returned)
$\sin^{-1} x$	$[-1,1]$	$\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$
$\cos^{-1} x$	$[-1,1]$	$[0,\pi]$
$\tan^{-1} x$	$\mathbb{R}$	$\left(-\tfrac{\pi}{2},\tfrac{\pi}{2}\right)$
$\cot^{-1} x$	$\mathbb{R}$	$(0,\pi)$
$\sec^{-1} x$	$(-\infty,-1]\cup[1,\infty)$	$[0,\pi]\setminus\{\tfrac{\pi}{2}\}$
$\csc^{-1} x$	$(-\infty,-1]\cup[1,\infty)$	$\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]\setminus\{0\}$

List - I	List - II
A) Domain of $\sin^{-1} x$	i) $\left(\dfrac{-\pi}{2}, \dfrac{\pi}{2}\right)$
B) Range of $\tan^{-1} x$	ii) $[0, \pi]$
C) Range of $\cos^{-1} x$	iii) $[-1, 1]$

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 understanding — Principal Value Domain

Principal Value Domain (Principal Branch)

The idea: restrict to one clean interval

Why the intervals differ …

Method: Evaluating a difference of inverse trig values

Steps

Common Mistakes

Concept and Intuition

Step-by-step solution

Q.tan⁡−13−sec⁡−1(−2)\tan^{-1} \sqrt{3} - \sec^{-1}(-2)tan−13​−sec−1(−2) is equal to (A) π\piπ (B) −π3-\frac{\pi}{3}−3π​ (C) π3\frac{\pi}{3}3π​ (D) 2π3\frac{2\pi}{3}32π​

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

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