Question

\( \tan^{-1} \sqrt{3} - \sec^{-1}(-2) \) is equal to \( \frac{\pi}{3} \).

Hint:

When working with inverse trigonometric functions, make sure to consider the principal values and the range of the functions.

Answer 1

Step 1: Evaluating \( \tan^{-1} \sqrt{3} \).
We know that: \[ \tan \left( \frac{\pi}{3} \right) = \sqrt{3} \] Therefore, \[ \tan^{-1} \sqrt{3} = \frac{\pi}{3} \] Step 2: Evaluating \( \sec^{-1}(-2) \).
The secant function \( \sec \theta = -2 \) corresponds to an angle where \( \cos \theta = -\frac{1}{2} \). The principal value of \( \sec^{-1}(-2) \) is \( \frac{2\pi}{3} \), since \( \sec^{-1} \) gives values in the range \( [0, \pi] \), and \( \cos \frac{2\pi}{3} = -\frac{1}{2} \). Step 3: Combining the results.
Now, we combine the results: \[ \tan^{-1} \sqrt{3} - \sec^{-1}(-2) = \frac{\pi}{3} - \frac{2\pi}{3} = -\frac{\pi}{3} \] Conclusion:
Thus, the statement is False, as the correct result is \( -\frac{\pi}{3} \), not \( \frac{\pi}{3} \).