Question

The value of $\tan^{-1}(\sqrt{3}) + \sec^{-1}(-2) - \sin^{-1}\left(-\frac{1}{2}\right)$ is

• A. $\frac{\pi}{2}$
• B. $\frac{7\pi}{6}$
• C. $\frac{2\pi}{3}$
• D. $\frac{5\pi}{6}$

Hint:

Remember principal value rules:

• \(\sin^{-1}(-x) = -\sin^{-1}(x)\)
• \(\tan^{-1}(-x) = -\tan^{-1}(x)\)
• \(\sec^{-1}(-x) = \pi - \sec^{-1}(x)\)

These shortcuts help evaluate inverse trigonometric expressions quickly.

Answer 1

The Correct Option is B

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

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

Step 1: {Evaluate \( \tan^{-1}(\sqrt{3}) \).} Since \[ \tan\left(\frac{\pi}{3}\right) = \sqrt{3} \] and \(\frac{\pi}{3}\) lies in the principal range, \[ \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \]
Step 2: {Evaluate \( \sec^{-1}(-2) \).} Let \[ \sec^{-1}(-2) = \theta \quad \Rightarrow \quad \sec\theta = -2 \] \[ \cos\theta = -\frac{1}{2} \] In the principal range \([0,\pi]\), this occurs at: \[ \theta = \frac{2\pi}{3} \] Thus, \[ \sec^{-1}(-2) = \frac{2\pi}{3} \]
Step 3: {Evaluate \( \sin^{-1}\left(-\frac{1}{2}\right) \).} We know \[ \sin\left(\frac{\pi}{6}\right) = \frac{1}{2} \] Using the property \( \sin^{-1}(-x) = -\sin^{-1}(x) \): \[ \sin^{-1}\left(-\frac{1}{2}\right) = -\frac{\pi}{6} \]
Step 4: {Add the values.} \[ \frac{\pi}{3} + \frac{2\pi}{3} - \left(-\frac{\pi}{6}\right) \] \[ = \pi + \frac{\pi}{6} \] \[ = \frac{7\pi}{6} \]
Step 5: {Conclusion.} \[ \tan^{-1}(\sqrt{3}) + \sec^{-1}(-2) - \sin^{-1}\left(-\frac{1}{2}\right) = \frac{7\pi}{6} \]