Evaluating tan[1/2 sec^-1(2/√3)]

This problem involves evaluating a trigonometric expression with an inverse trigonometric function. We need to find the value of \(\tan \left[ \frac{1}{2} \sec^{-1} \left( \frac{2}{\sqrt{3}} \right) \right]\). We will break this down step-by-step.

Step-by-Step Calculation

1. Identify the inverse function: First, let's focus on the inverse function part: \(\sec^{-1} \left( \frac{2}{\sqrt{3}} \right)\). Let this angle be \(\theta\). So, \(\theta = \sec^{-1} \left( \frac{2}{\sqrt{3}} \right)\).

   By definition of the inverse secant function, this means \(\sec \theta = \frac{2}{\sqrt{3}}\).

2. Find the angle \(\theta\): We know that \(\sec \theta = \frac{1}{\cos \theta}\). Therefore, \(\cos \theta = \frac{1}{\sec \theta} = \frac{1}{2/\sqrt{3}} = \frac{\sqrt{3}}{2}\).

   We need to find the angle \(\theta\) (in the principal value range of \(\sec^{-1}\), which is \([0, \pi]\) excluding \(\frac{\pi}{2}\)) for which \(\cos \theta = \frac{\sqrt{3}}{2}\). This angle is \(\theta = \frac{\pi}{6}\).

3. Substitute back into the expression: Now we substitute this value of \(\theta\) back into the original expression:

   \(\tan \left[ \frac{1}{2} \sec^{-1} \left( \frac{2}{\sqrt{3}} \right) \right] = \tan \left( \frac{1}{2} \theta \right) = \tan \left( \frac{1}{2} \cdot \frac{\pi}{6} \right)\)

   This simplifies to \(\tan \left( \frac{\pi}{12} \right)\).

4. Evaluate tan(π/12): We need to find the value of \(\tan \left( \frac{\pi}{12} \right)\). We can do this using the tangent half-angle formula or the tangent subtraction formula.

   Method 1: Using Tangent Half-Angle Formula

   The formula is \(\tan \left( \frac{x}{2} \right) = \frac{1 - \cos x}{\sin x}\). Here, \(x = \frac{\pi}{6}\).

   We know \(\cos \left( \frac{\pi}{6} \right) = \frac{\sqrt{3}}{2}\) and \(\sin \left( \frac{\pi}{6} \right) = \frac{1}{2}\).

   Substituting these values:

   \(\tan \left( \frac{\pi}{12} \right) = \tan \left( \frac{\pi/6}{2} \right) = \frac{1 - \cos(\pi/6)}{\sin(\pi/6)} = \frac{1 - \frac{\sqrt{3}}{2}}{\frac{1}{2}}\)

   Simplify the fraction:

   \(\frac{\frac{2 - \sqrt{3}}{2}}{\frac{1}{2}} = 2 - \sqrt{3}\)

   Method 2: Using Tangent Subtraction Formula

   We can write \(\frac{\pi}{12}\) as \(\frac{\pi}{4} - \frac{\pi}{6}\). The formula is \(\tan(A - B) = \frac{\tan A - \tan B}{1 + \tan A \tan B}\).

   Here, \(A = \frac{\pi}{4}\) and \(B = \frac{\pi}{6}\). We know \(\tan \left( \frac{\pi}{4} \right) = 1\) and \(\tan \left( \frac{\pi}{6} \right) = \frac{1}{\sqrt{3}}\).

   Substituting these values:

   \(\tan \left( \frac{\pi}{12} \right) = \tan \left( \frac{\pi}{4} - \frac{\pi}{6} \right) = \frac{\tan(\pi/4) - \tan(\pi/6)}{1 + \tan(\pi/4) \tan(\pi/6)}\) \(= \frac{1 - \frac{1}{\sqrt{3}}}{1 + 1 \cdot \frac{1}{\sqrt{3}}} = \frac{\frac{\sqrt{3} - 1}{\sqrt{3}}}{\frac{\sqrt{3} + 1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1}\)

   To simplify, rationalize the denominator:

   \(\frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} = \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - 1^2} = \frac{3 - 2\sqrt{3} + 1}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}\)

5. Final Answer: Both methods show that the value of the expression is \(2 - \sqrt{3}\).

Therefore, the value of \(\tan \left[ \frac{1}{2} \sec^{-1} \left( \frac{2}{\sqrt{3}} \right) \right]\) is \(2 - \sqrt{3}\).