Given that $\csc(x) = 9$ for $90^\circ < x < 180^\circ$, we can use this information to find the values of $\sin\left(\frac{x}{2}\right)$, $\cos\left(\frac{x}{2}\right)$, and $\tan\left(\frac{x}{2}\right)$.

Step 1: Find $\sin(x)$

Since $\csc(x) = 9$, we know: $\csc(x) = \frac{1}{\sin(x)} \quad \text{so} \quad \sin(x) = \frac{1}{\csc(x)} = \frac{1}{9}.$ Thus, $\sin(x) = \frac{1}{9}$.

Step 2: Determine $x$

We are given that $90^\circ < x < 180^\circ$. In this interval, $x$ is in the second quadrant, where the sine function is positive.

Step 3: Apply the Half-Angle Formulas

Half-Angle Formula for $\sin\left(\frac{x}{2}\right)$:

$\sin\left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - \cos(x)}{2}}.$ We'll need to find $\cos(x)$ to proceed.

Find $\cos(x)$:

Using the Pythagorean identity $\sin^2(x) + \cos^2(x) = 1$, we can find $\cos(x)$: $\sin(x) = \frac{1}{9}, \quad \text{so} \quad \sin^2(x) = \left(\frac{1}{9}\right)^2 = \frac{1}{81}.$ Now, $\cos^2(x) = 1 - \sin^2(x) = 1 - \frac{1}{81} = \frac{80}{81},$ so $\cos(x) = -\sqrt{\frac{80}{81}} = -\frac{\sqrt{80}}{9} = -\frac{4\sqrt{5}}{9},$ since $x$ is in the second quadrant where cosine is negative.

Find $\sin\left(\frac{x}{2}\right)$:

Now, use the half-angle formula for sine: $\sin\left(\frac{x}{2}\right) = \pm \sqrt{\frac{1 - \cos(x)}{2}} = \pm \sqrt{\frac{1 + \frac{4\sqrt{5}}{9}}{2}}.$ We'll determine the sign based on the quadrant once we simplify.

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Related Questions:

1. How can we determine in which quadrant $\frac{x}{2}$ lies?
2. What are the half-angle formulas for other trigonometric functions like tangent?
3. How can the Pythagorean identity be derived?
4. How do trigonometric functions behave in different quadrants?
5. How can we verify the solutions using a calculator or graph?

Tip: Always check the quadrant of the angle to determine the correct sign of the trigonometric function.