First, we find the principal value of each inverse trigonometric function:
\(\tan^{-1}(1)\): The principal value is \(\frac{\pi}{4}\) because \(\tan(\frac{\pi}{4}) = 1\) and \(-\frac{\pi}{2} < \frac{\pi}{4} < \frac{\pi}{2}\).
\(\cos^{-1}(-\frac{1}{2})\): The principal value is \(\frac{2\pi}{3}\) because \(\cos(\frac{2\pi}{3}) = -\frac{1}{2}\) and \(0 \le \frac{2\pi}{3} \le \pi\).
\(\sin^{-1}(-\frac{1}{\sqrt{2}})\): The principal value is \(-\frac{\pi}{4}\) because \(\sin(-\frac{\pi}{4}) = -\frac{1}{\sqrt{2}}\) and \(-\frac{\pi}{2} \le -\frac{\pi}{4} \le \frac{\pi}{2}\).
Now, we add the principal values:
\(\tan^{-1}(1) + \cos^{-1}(-\frac{1}{2}) + \sin^{-1}(-\frac{1}{\sqrt{2}}) = \frac{\pi}{4} + \frac{2\pi}{3} - \frac{\pi}{4}\)
\(= \frac{2\pi}{3}\)
Correct Answer: 2π/3
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