Q.The value of the expression $2\sec^{-1}2+\sin^{-1}\frac{1}{2}$ is
(A) $\frac{\pi}{6}$
(B) $\frac{5\pi}{6}$
(C) $\frac{7\pi}{6}$
(D) $1$

Concept: Inverse trigonometric functions — principal value ranges and standard angle evaluation.

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

Since $\sec \frac{\pi}{3} = 2$ and the principal value range of $\sec^{-1}$ is $[0,\pi] \setminus \{\frac{\pi}{2}\}$, we have $\sec^{-1}2 = \frac{\pi}{3}$.

Step 2: Evaluate $\sin^{-1}\frac{1}{2}$.

Since $\sin \frac{\pi}{6} = \frac{1}{2}$ and the principal value range of $\sin^{-1}$ is $[-\frac{\pi}{2}, \frac{\pi}{2}]$, we have $\sin^{-1}\frac{1}{2} = \frac{\pi}{6}$.

Step 3: Compute the expression. …

Concept and Intuition

When you see an expression like $2\sec^{-1}2 + \sin^{-1}\frac{1}{2}$, the natural instinct is to reach for a calculator or try to recall the exact values. But the real skill here is knowing the principal value ranges of inverse trigonometric functions. Without that, you might pick the wrong angle.

For $\sec^{-1}x$, the principal value range is $[0, \pi]$ excluding $\frac{\pi}{2}$. For $\sin^{-1}x$, it's $[-\frac{\pi}{2}, \frac{\pi}{2}]$. These ranges are not arbitrary — they ensure that each inverse function gives a unique output. The exam expects you to know these cold.

Let’s work through it.

1. Evaluate $\sec^{-1}2$

   We need an angle $\theta$ such that $\sec\theta = 2$ and $\theta$ lies in the principal range of $\sec^{-1}$, which is $[0, \pi]$ with $\theta \neq \frac{\pi}{2}$.

   Since $\sec\theta = \frac{1}{\cos\theta}$, we have $\cos\theta = \frac{1}{2}$. The angles in $[0, \pi]$ where $\cos\theta = \frac{1}{2}$ are $\theta = \frac{\pi}{3}$ and $\theta = \frac{5\pi}{3}$ — but $\frac{5\pi}{3}$ is outside $[0, \pi]$. So the only candidate is $\theta = \frac{\pi}{3}$.

   Tip

   A quick check: $\sec\frac{\pi}{3} = 2$, and $\frac{\pi}{3}$ lies in $[0, \pi]$, so it's valid.

   Hence, $\sec^{-1}2 = \frac{\pi}{3}$.

2. Multiply by 2

   The expression has $2\sec^{-1}2$, so:

3. Evaluate $\sin^{-1}\frac{1}{2}$

   We need an angle $\phi$ such that $\sin\phi = \frac{1}{2}$ and $\phi$ lies in the principal range of $\sin^{-1}$, which is $[-\frac{\pi}{2}, \frac{\pi}{2}]$. …

Method: Evaluating a combination of inverse-trig functions at standard values

For expressions like $2\sec^{-1}2+\sin^{-1}\tfrac12$, evaluate each inverse function separately as a standard-angle principal value, then combine.

Steps

Step 1: Rewrite each reciprocal inverse in terms of a basic one.

Convert $\sec^{-1}$ / $\csc^{-1}$ into cosine/sine problems: $\sec^{-1}m$ means the angle whose cosine is $\tfrac1m$, taken in the $\sec^{-1}$ range $[0,\pi]\setminus\{\tfrac{\pi}{2}\}$. So $\sec^{-1}2$ is the angle with $\cos\theta=\tfrac12$.

Step 2: Read off each standard angle in its own principal range. …

Common Mistakes

Mistake 1: Forgetting the coefficient $2$ on $\sec^{-1}2$.

Why it's wrong: dropping it gives $\tfrac{\pi}{3}+\tfrac{\pi}{6}=\tfrac{\pi}{2}$, not the required value. Correct approach: $2\sec^{-1}2=2\cdot\tfrac{\pi}{3}=\tfrac{2\pi}{3}$, then add $\tfrac{\pi}{6}$ to get $\tfrac{5\pi}{6}$.

Mistake 2: Taking $\sin^{-1}\tfrac12=\tfrac{5\pi}{6}$.

Why it's wrong: although $\sin\tfrac{5\pi}{6}=\tfrac12$, that angle is outside the $\sin^{-1}$ range $\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$. Correct approach: $\sin^{-1}\tfrac12=\tfrac{\pi}{6}$. …