Question

The value of \(\tan^{-1}(√3)-\sec^{-1}(\dfrac{2}{√3})\) is ?

• A. \(\dfrac{2π}{3}\)
• B. \(\dfrac{π}{4}\)
• C. \(\dfrac{π}{2}\)
• D. \(\dfrac{π}{6}\)
• E. \(\dfrac{π}{3}\)

Answer 1

The Correct Option is D

Step 1: Evaluate \(\tan^{-1}(\sqrt{3})\). Recall that \(\tan\left(\frac{\pi}{3}\right) = \sqrt{3}\), so: \[ \tan^{-1}(\sqrt{3}) = \frac{\pi}{3} \]

Step 2: Evaluate \(\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)\). Let \(\theta = \sec^{-1}\left(\frac{2}{\sqrt{3}}\right)\), then \(\sec \theta = \frac{2}{\sqrt{3}}\). This implies \(\cos \theta = \frac{\sqrt{3}}{2}\), so: \[ \theta = \frac{\pi}{6} \]

Step 3: Subtract the two results: \[ \tan^{-1}(\sqrt{3}) - \sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \frac{\pi}{3} - \frac{\pi}{6} = \frac{\pi}{6} \]

Conclusion: The correct answer is \(\boxed{D}\) \(\left(\frac{\pi}{6}\right)\).

Answer 2

Evaluating each term:

• \(\tan^{-1}(\sqrt{3})\): The angle whose tangent is \(\sqrt{3}\) is \(\frac{\pi}{3}\). However, the range of the principal value of \(\tan^{-1}(x)\) is \((- \frac{\pi}{2}, \frac{\pi}{2})\). Since \(\sqrt{3}\) is positive, \(\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}\).
• \(\sec^{-1}\left(\frac{2}{\sqrt{3}}\right)\): The secant function is the reciprocal of the cosine function. So, \(\sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \cos^{-1}\left(\frac{\sqrt{3}}{2}\right)\). The angle whose cosine is \(\frac{\sqrt{3}}{2}\) is \(\frac{\pi}{6}\). The range of the principal value of \(\sec^{-1}(x)\) is \([0, \pi]\) excluding \(\frac{\pi}{2}\). Since \(\frac{2}{\sqrt{3}} > 1\), we have \(\sec^{-1}\left(\frac{2}{\sqrt{3}}\right) = \frac{\pi}{6}\).

Therefore:

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