Q7:

Find the principal values of the following: $\sec^{-1} (\frac{2}{\sqrt{3}})$

Step 1: Theoretical Foundation of Principal Value Branch

The inverse secant function, denoted as $\sec^{-1}(x)$, is defined such that its principal value branch lies within the interval $[0, \pi] \setminus \{\frac{\pi}{2}\}$. For a given value $x$, the principal value $y$ must satisfy the condition:

$\sec(y) = x$, where $y \in [0, \pi], y \neq \frac{\pi}{2}$ [Per the definition of the range of the inverse secant function].

Step 2: Mathematical Setup and Transformation

Given the equation: $\sec^{-1} \left( \frac{2}{\sqrt{3}} \right) = y$

By the definition of the inverse trigonometric function, we convert the expression into its equivalent trigonometric form:

$\sec(y) = \frac{2}{\sqrt{3}}$

Since $\sec(y) = \frac{1}{\cos(y)}$, we apply the reciprocal identity:

$\cos(y) = \frac{1}{\sec(y)} = \frac{1}{\frac{2}{\sqrt{3}}} = \frac{\sqrt{3}}{2}$

Step 3: Determining the Principal Value

To solve $\cos(y) = \frac{\sqrt{3}}{2}$ for $y$ within the principal interval $[0, \pi]$:

• We know from standard trigonometric table values that $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$. [By the unit circle definition of cosine].
• Since $\frac{\pi}{6}$ lies within the interval $[0, \pi]$, it satisfies the condition of the principal value branch.

Step 4: Geometric Visualization

Step 5: Verification of Constraints

Check: Does $y = \frac{\pi}{6}$ satisfy $y \in [0, \pi] \setminus \{\frac{\pi}{2}\}$?

Since $0 \leq \frac{\pi}{6} \leq \pi$ and $\frac{\pi}{6} \neq \frac{\pi}{2}$, the value is valid. [Per the domain and range constraints of $\sec^{-1}(x)$].

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