Evaluating Inverse Trigonometric Expressions

The question requires evaluating the sum $\cos^{-1}\left(\frac{1}{\sqrt{2}}\right) + \sec^{-1}(-\sqrt{2})$. We evaluate each term separately.

Evaluating $\cos^{-1}\left(\frac{1}{\sqrt{2}}\right)$

• The inverse cosine function, $\cos^{-1}(x)$, returns an angle $\theta$ in the range $[0, \pi]$ such that $\cos(\theta) = x$.
• We need to find $\theta$ where $\cos(\theta) = \frac{1}{\sqrt{2}}$ and $0 \le \theta \le \pi$.
• The angle that satisfies this condition is $\theta = \frac{\pi}{4}$.
• Therefore, $\cos^{-1}\left(\frac{1}{\sqrt{2}}\right) = \frac{\pi}{4}$.

Evaluating $\sec^{-1}(-\sqrt{2})$

• The inverse secant function, $\sec^{-1}(x)$, is typically defined such that its range is $[0, \pi]$, excluding $\frac{\pi}{2}$.
• We know that $\sec^{-1}(x) = \cos^{-1}\left(\frac{1}{x}\right)$ for $|x| \ge 1$.
• So, $\sec^{-1}(-\sqrt{2}) = \cos^{-1}\left(\frac{1}{-\sqrt{2}}\right) = \cos^{-1}\left(-\frac{1}{\sqrt{2}}\right)$.
• We need to find an angle $\phi$ in the range $[0, \pi]$ such that $\cos(\phi) = -\frac{1}{\sqrt{2}}$.
• Since the cosine value is negative, the angle must be in the second quadrant (as the range is $[0, \pi]$).
• The reference angle for which $\cos(\text{ref}) = \frac{1}{\sqrt{2}}$ is $\frac{\pi}{4}$.
• Therefore, the angle in the second quadrant is $\phi = \pi - \frac{\pi}{4} = \frac{3\pi}{4}$.
• Thus, $\sec^{-1}(-\sqrt{2}) = \frac{3\pi}{4}$.

Calculating the Final Sum

• Now, we add the results of the two inverse trigonometric functions: $ \cos^{-1}\left(\frac{1}{\sqrt{2}}\right) + \sec^{-1}(-\sqrt{2}) = \frac{\pi}{4} + \frac{3\pi}{4} $
• Combine the terms: $ \frac{\pi + 3\pi}{4} = \frac{4\pi}{4} $
• Simplify the expression: $ \frac{4\pi}{4} = \pi $

Conclusion

The value of the expression $\cos^{-1}\left(\frac{1}{\sqrt{2}}\right) + \sec^{-1}(-\sqrt{2})$ is $\pi$.