To solve this expression, we need to simplify $\csc \left( \arccos \left( \frac{9}{x} \right) \right)$.

Let's go through the steps:

1. Define the Angle: Let $\theta = \arccos \left( \frac{9}{x} \right)$. This means that $\cos(\theta) = \frac{9}{x}$.

2. Find $\sin(\theta)$: To express $\csc(\theta) = \frac{1}{\sin(\theta)}$, we need $\sin(\theta)$.

   • Using the Pythagorean identity, $\sin^2(\theta) + \cos^2(\theta) = 1$.
   • Substitute $\cos(\theta) = \frac{9}{x}$:
     $\sin^2(\theta) + \left( \frac{9}{x} \right)^2 = 1$ $\sin^2(\theta) + \frac{81}{x^2} = 1$ $\sin^2(\theta) = 1 - \frac{81}{x^2}$ $\sin(\theta) = \sqrt{1 - \frac{81}{x^2}}$

3. Find $\csc(\theta)$: Now that $\sin(\theta) = \sqrt{1 - \frac{81}{x^2}}$, $\csc(\theta) = \frac{1}{\sin(\theta)} = \frac{1}{\sqrt{1 - \frac{81}{x^2}}}$

Thus, the equivalent algebraic expression is: $\csc \left( \arccos \left( \frac{9}{x} \right) \right) = \frac{1}{\sqrt{1 - \frac{81}{x^2}}}$

Would you like more details or have any questions?

Here are five related questions:

1. How would this expression change if it were $\sec(\arccos(\frac{9}{x}))$ instead of $\csc$?
2. What if we had $\sin(\arccos(\frac{9}{x}))$ instead of $\csc$?
3. How does understanding trigonometric identities help in simplifying such expressions?
4. Why do we use the Pythagorean identity in this simplification?
5. Could we encounter similar expressions with other inverse trigonometric functions like $\arcsin$ or $\arctan$?

Tip: When dealing with inverse trigonometric functions, try assigning a variable to the angle, then use trigonometric identities to find equivalent expressions.