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

The expression $\sec^{-1} \left( -\sqrt{2} \right) - \tan^{-1} \left( \frac{1}{\sqrt{3}} \right)$ is evaluated. For $\sec^{-1} \left( -\sqrt{2} \right)$, we seek $\theta$ where $\sec \theta = -\sqrt{2}$. This yields $\theta = \frac{3\pi}{4}$. For $\tan^{-1} \left( \frac{1}{\sqrt{3}} \right)$, we seek $\alpha$ where $\tan \alpha = \frac{1}{\sqrt{3}}$. This yields $\alpha = \frac{\pi}{6}$. Substituting these values, the expression becomes: \[ \frac{3\pi}{4} - \frac{\pi}{6} \] To perform the subtraction, we find a common denominator of 12: \[ \frac{3\pi}{4} = \frac{9\pi}{12}, \quad \frac{\pi}{6} = \frac{2\pi}{12} \] Subtracting the fractions: \[ \frac{9\pi}{12} - \frac{2\pi}{12} = \frac{7\pi}{12} \] The simplified result is $\frac{7\pi}{12}$.