Find the principal values of the following :
`sec^(-1) (2/(sqrt(3)))`

Answer

To find the principal value of \( \sec^{-1} \left( \frac{2}{\sqrt{3}} \right) \), we will follow these steps:

### Step 1: Understand the Definition
The function \( \sec^{-1}(x) \) is the inverse of the secant function. The domain of \( \sec^{-1}(x) \) is \( (-\infty, -1] \cup [1, \infty) \) and the range is \( [0, \frac{\pi}{2}) \cup (\frac{\pi}{2}, \pi] \).

### Step 2: Set Up the Equation
Let \( y = \sec^{-1} \left( \frac{2}{\sqrt{3}} \right) \). By the definition of the inverse secant function, we have:
\[ \sec(y) = \frac{2}{\sqrt{3}} \]

### Step 3: Convert to Cosine
Since \( \sec(y) = \frac{1}{\cos(y)} \), we can write:
\[ \frac{1}{\cos(y)} = \frac{2}{\sqrt{3}} \]
This implies:
\[ \cos(y) = \frac{\sqrt{3}}{2} \]

### Step 4: Find the Angle
Now, we need to find the angle \( y \) such that:
\[ \cos(y) = \frac{\sqrt{3}}{2} \]
The angle \( y \) that satisfies this equation in the range of \( [0, \pi] \) is:
\[ y = \frac{\pi}{6} \]

### Step 5: Conclusion
Thus, the principal value of \( \sec^{-1} \left( \frac{2}{\sqrt{3}} \right) \) is:
\[ \sec^{-1} \left( \frac{2}{\sqrt{3}} \right) = \frac{\pi}{6} \]

### Final Answer:
\[ \sec^{-1} \left( \frac{2}{\sqrt{3}} \right) = \frac{\pi}{6} \]

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