If $\cos 27^\circ = \frac{p}{q}$, then find the value of $\csc 27^\circ - \cos 63^\circ$.

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**Solution:**

1. Given $\cos 27^\circ = \frac{p}{q}$.

2. Recall the identity: $\csc A = \frac{1}{\sin A}$ and $\cos (90^\circ - A) = \sin A$.

3. Note that $\cos 63^\circ = \cos (90^\circ - 27^\circ) = \sin 27^\circ$.

4. Therefore,
   $$\csc 27^\circ - \cos 63^\circ = \frac{1}{\sin 27^\circ} - \sin 27^\circ = \frac{1 - \sin^2 27^\circ}{\sin 27^\circ}$$

5. Using the Pythagorean identity, $1 - \sin^2 27^\circ = \cos^2 27^\circ$.

6. Substitute $\cos 27^\circ = \frac{p}{q}$:
   $$\cos^2 27^\circ = \left( \frac{p}{q} ight)^2 = \frac{p^2}{q^2}$$

7. Also, $\sin^2 27^\circ = 1 - \cos^2 27^\circ = 1 - \frac{p^2}{q^2} = \frac{q^2 - p^2}{q^2}$.

8. Hence,
   $$\sin 27^\circ = \frac{\sqrt{q^2 - p^2}}{q}$$

9. Substitute these into the expression:
   $$\csc 27^\circ - \cos 63^\circ = \frac{\frac{p^2}{q^2}}{\frac{\sqrt{q^2 - p^2}}{q}} = \frac{p^2}{q^2} 	imes \frac{q}{\sqrt{q^2 - p^2}} = \frac{p^2}{q \sqrt{q^2 - p^2}}$$

**Answer:** $\boxed{\frac{p^2}{q \sqrt{q^2 - p^2}}}$

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