Question

Find the principal value of \( \tan^{-1}(\sqrt{3}) - \sec^{-1}(-2) \).

Hint:

Remember the principal value ranges: \(\tan^{-1}x \in \left(-\frac{\pi}{2},\frac{\pi}{2}\right)\) and \(\sec^{-1}x \in [0,\pi]\). Also recall standard values such as \(\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}\) and \(\sec^{-1}(-2) = \frac{2\pi}{3}\).

Answer 1

Concept: To evaluate expressions involving inverse trigonometric functions, we use their principal value ranges:

• \( \tan^{-1}x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right) \)
• \( \sec^{-1}x \in [0,\pi], \; x\neq (-1,1) \)

We use standard trigonometric values and identities to determine the angles.
Step 1: Evaluate \( \tan^{-1}(\sqrt{3}) \).
We know that \[ \tan \frac{\pi}{3} = \sqrt{3} \] Since \( \frac{\pi}{3} \) lies within the principal value range of \( \tan^{-1}x \), \[ \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \]
Step 2: Evaluate \( \sec^{-1}(-2) \).
Let \[ \sec^{-1}(-2) = \theta \] Then \[ \sec\theta = -2 \] \[ \cos\theta = -\frac{1}{2} \] In the principal value range \( [0,\pi] \), the angle whose cosine is \( -\frac{1}{2} \) is \[ \theta = \frac{2\pi}{3} \] Thus, \[ \sec^{-1}(-2) = \frac{2\pi}{3} \]
Step 3: Compute the required expression.
\[ \tan^{-1}(\sqrt{3}) - \sec^{-1}(-2) \] \[ = \frac{\pi}{3} - \frac{2\pi}{3} \] \[ = -\frac{\pi}{3} \]
Step 4: Final Answer.
\[ \tan^{-1}(\sqrt{3}) - \sec^{-1}(-2) = -\frac{\pi}{3} \]