Let's break down the problems from the image you uploaded:

Problem 91: $\sin^{-1}(0)$

• The inverse sine of $0$ asks for the angle whose sine value is 0.
• $\sin^{-1}(0) = 0$, because the sine of $0$ radians is 0.

Problem 94: $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right)$

• The inverse cosine of $\frac{\sqrt{3}}{2}$ asks for the angle whose cosine value is $\frac{\sqrt{3}}{2}$.
• The angle whose cosine is $\frac{\sqrt{3}}{2}$ is $\frac{\pi}{6}$ radians, or 30 degrees.

Thus, $\cos^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{6}$.

Problem 97: $\tan^{-1}(\sqrt{3})$

• The inverse tangent of $\sqrt{3}$ asks for the angle whose tangent value is $\sqrt{3}$.
• The angle whose tangent is $\sqrt{3}$ is $\frac{\pi}{3}$ radians, or 60 degrees.

Thus, $\tan^{-1}(\sqrt{3}) = \frac{\pi}{3}$.

Problem 90: $\tan^{-1}\left(\frac{\sqrt{3}}{3}\right)$

• The inverse tangent of $\frac{\sqrt{3}}{3}$ asks for the angle whose tangent value is $\frac{\sqrt{3}}{3}$.
• The angle whose tangent is $\frac{\sqrt{3}}{3}$ is $\frac{\pi}{6}$ radians, or 30 degrees.

Thus, $\tan^{-1}\left(\frac{\sqrt{3}}{3}\right) = \frac{\pi}{6}$.

Problem 93: $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right)$

• The inverse sine of $\frac{\sqrt{3}}{2}$ asks for the angle whose sine value is $\frac{\sqrt{3}}{2}$.
• The angle whose sine is $\frac{\sqrt{3}}{2}$ is $\frac{\pi}{3}$ radians, or 60 degrees.

Thus, $\sin^{-1}\left(\frac{\sqrt{3}}{2}\right) = \frac{\pi}{3}$.

Would you like any additional details or explanations on these solutions?

Here are five related questions for practice:

1. What is $\sin^{-1}(1)$?
2. What is $\cos^{-1}(-1)$?
3. Solve $\tan^{-1}(-1)$.
4. What is $\sin^{-1}\left(\frac{1}{2}\right)$?
5. Solve $\cos^{-1}(0)$.

Tip: When working with inverse trigonometric functions, always consider the principal values or the specific range of angles where the function is defined.