Let’s evaluate each expression step by step:

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(a) $\sin^{-1}(1)$

The inverse sine function ($\sin^{-1}(x)$) gives the angle $\theta$ such that $\sin(\theta) = x$, where $-\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2}$.

For $\sin^{-1}(1)$: $\sin(\theta) = 1$ The angle where this occurs is: $\theta = \frac{\pi}{2}$

Answer: $\frac{\pi}{2}$

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(b) $\cos^{-1}(0)$

The inverse cosine function ($\cos^{-1}(x)$) gives the angle $\theta$ such that $\cos(\theta) = x$, where $0 \leq \theta \leq \pi$.

For $\cos^{-1}(0)$: $\cos(\theta) = 0$ The angle where this occurs is: $\theta = \frac{\pi}{2}$

Answer: $\frac{\pi}{2}$

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(c) $\tan^{-1}(\sqrt{3})$

The inverse tangent function ($\tan^{-1}(x)$) gives the angle $\theta$ such that $\tan(\theta) = x$, where $-\frac{\pi}{2} < \theta < \frac{\pi}{2}$.

For $\tan^{-1}(\sqrt{3})$: $\tan(\theta) = \sqrt{3}$ The angle where this occurs is: $\theta = \frac{\pi}{3}$

Answer: $\frac{\pi}{3}$

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Final Answers:

(a) $\frac{\pi}{2}$
(b) $\frac{\pi}{2}$
(c) $\frac{\pi}{3}$

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Do you want more details on the solution or explanations? 😊

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5 Related Questions:

1. What is the domain and range of the inverse sine function ($\sin^{-1}(x)$)?
2. How do you find the principal value of inverse trigonometric functions?
3. What is the value of $\tan^{-1}(-\sqrt{3})$ in radians?
4. Why is the range of $\cos^{-1}(x)$ restricted to $[0, \pi]$?
5. How are inverse trigonometric functions used to solve trigonometric equations?

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One Tip:

To work comfortably with inverse trigonometric functions, remember their domain and range restrictions—they ensure the functions are well-defined and one-to-one.