Question

\[ \sec^{-1} \left( -\sqrt{2} \right) - \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) \] is equal to:

• A. $\frac{11\pi}{12}$
• B. $\frac{5\pi}{12}$
• C. $-\frac{5\pi}{12}$
• D. $\frac{7\pi}{12}$

Hint:

When dealing with inverse trigonometric functions, recall their principal values and how they relate to angles on the unit circle.

Answer 1

The Correct Option is C

We are given the expression $\sec^{-1} \left( -\sqrt{2} \right) - \tan^{-1} \left( \frac{1}{\sqrt{3}} \right)$. We know that: - $\sec^{-1} \left( -\sqrt{2} \right)$ corresponds to the angle $\theta$ such that $\sec \theta = -\sqrt{2}$. The value of $\theta$ is $\frac{3\pi}{4}$. - $\tan^{-1} \left( \frac{1}{\sqrt{3}} \right)$ corresponds to the angle $\alpha$ such that $\tan \alpha = \frac{1}{\sqrt{3}}$, which gives $\alpha = \frac{\pi}{6}$. Thus, the expression becomes: \[ \sec^{-1} \left( -\sqrt{2} \right) - \tan^{-1} \left( \frac{1}{\sqrt{3}} \right) = \frac{3\pi}{4} - \frac{\pi}{6}. \] To simplify, we need a common denominator: \[ \frac{3\pi}{4} = \frac{9\pi}{12}, \quad \frac{\pi}{6} = \frac{2\pi}{12}. \] Now, subtract: \[ \frac{9\pi}{12} - \frac{2\pi}{12} = \frac{7\pi}{12}. \] Thus, the final answer is $\frac{7\pi}{12}$.