The value of `2sec^(-1) + 2 sin^(-1)(1/2)` is

Options

A. `(pi)/(6)`
B. `(5pi)/(6)`
C. `(7pi)/(6)`
D. `1`

Explanation

To solve the expression \( 2 \sec^{-1}(2) + 2 \sin^{-1}\left(\frac{1}{2}\right) \), we will break it down step by step.

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

The value of \( \sec^{-1}(x) \) is the angle \( \theta \) such that \( \sec(\theta) = x \). Therefore, we need to find \( \theta \) such that:

\[ \sec(\theta) = 2 \]

This implies:

\[ \cos(\theta) = \frac{1}{2} \]

The angle \( \theta \) that satisfies \( \cos(\theta) = \frac{1}{2} \) is:

\[ \theta = \frac{\pi}{3} \]

Thus, we have:

\[ \sec^{-1}(2) = \frac{\pi}{3} \]

### Step 2: Evaluate \( \sin^{-1}\left(\frac{1}{2}\right) \)

Next, we evaluate \( \sin^{-1}\left(\frac{1}{2}\right) \). The value of \( \sin^{-1}(x) \) is the angle \( \phi \) such that \( \sin(\phi) = x \). Therefore, we need to find \( \phi \) such that:

\[ \sin(\phi) = \frac{1}{2} \]

The angle \( \phi \) that satisfies \( \sin(\phi) = \frac{1}{2} \) is:

\[ \phi = \frac{\pi}{6} \]

Thus, we have:

\[ \sin^{-1}\left(\frac{1}{2}\right) = \frac{\pi}{6} \]

### Step 3: Substitute back into the original expression

Now we can substitute these values back into the original expression:

\[ 2 \sec^{-1}(2) + 2 \sin^{-1}\left(\frac{1}{2}\right) = 2 \left(\frac{\pi}{3}\right) + 2 \left(\frac{\pi}{6}\right) \]

### Step 4: Simplify the expression

Calculating each term:

\[ 2 \left(\frac{\pi}{3}\right) = \frac{2\pi}{3} \]
\[ 2 \left(\frac{\pi}{6}\right) = \frac{\pi}{3} \]

Now, we add these two results together:

\[ \frac{2\pi}{3} + \frac{\pi}{3} = \frac{2\pi + \pi}{3} = \frac{3\pi}{3} = \pi \]

### Final Result

Thus, the value of \( 2 \sec^{-1}(2) + 2 \sin^{-1}\left(\frac{1}{2}\right) \) is:

\[ \pi \]