### (a)

**Step 1 — Simplify each denominator using identities.**
$$\sec^2\theta - 1 = \tan^2\theta, \qquad \operatorname{cosec}^2\theta - 1 = \cot^2\theta$$

$$\text{LHS} = \frac{\sec^3\theta}{\tan^2\theta} + \frac{\operatorname{cosec}^3\theta}{\cot^2\theta}$$

**Step 2 — Convert everything to sine and cosine.**
$$= \frac{1/\cos^3\theta}{\sin^2\theta/\cos^2\theta} + \frac{1/\sin^3\theta}{\cos^2\theta/\sin^2\theta}$$

**Step 3 — Simplify each term.**
$$= \frac{1}{\cos^3\theta} \times \frac{\cos^2\theta}{\sin^2\theta} + \frac{1}{\sin^3\theta} \times \frac{\sin^2\theta}{\cos^2\theta}$$
$$= \frac{1}{\cos\theta\,\sin^2\theta} + \frac{1}{\sin\theta\,\cos^2\theta}$$

**Step 4 — Take LCM $= \sin^2\theta\cos^2\theta$.**
$$= \frac{\cos\theta + \sin\theta}{\sin^2\theta\,\cos^2\theta}$$

**Step 5 — Split the denominator.**
$$= \frac{1}{\sin\theta\cos\theta}\left(\frac{\cos\theta}{\sin\theta\cos\theta} + \frac{\sin\theta}{\sin\theta\cos\theta}\right)$$
$$= \sec\theta\operatorname{cosec}\theta\left(\frac{1}{\sin\theta} + \frac{1}{\cos\theta}\right)$$

**Step 6 — Write in terms of sec and cosec.**
$$= \sec\theta \cdot \operatorname{cosec}\theta\,(\operatorname{cosec}\theta + \sec\theta) = \text{RHS}$$

$$\textbf{Hence proved.}$$

---

### (b)

**Step 1 — Compute $p^2 - q^2$.**
$$p^2 - q^2 = \frac{\sec^2\alpha}{\operatorname{cosec}^2\beta} - \frac{\tan^2\alpha}{\operatorname{cosec}^2\beta} = \frac{\sec^2\alpha - \tan^2\alpha}{\operatorname{cosec}^2\beta}$$

**Step 2 — Use the identity $\sec^2\alpha - \tan^2\alpha = 1$.**
$$p^2 - q^2 = \frac{1}{\operatorname{cosec}^2\beta}$$

**Step 3 — Multiply both sides by $\sec^2\alpha$.**
$$(p^2 - q^2)\sec^2\alpha = \frac{\sec^2\alpha}{\operatorname{cosec}^2\beta}$$

**Step 4 — Recognise the RHS.**
$$\frac{\sec^2\alpha}{\operatorname{cosec}^2\beta} = \left(\frac{\sec\alpha}{\operatorname{cosec}\beta}\right)^2 = p^2$$

$$\therefore\ (p^2 - q^2)\sec^2\alpha = p^2 \qquad \textbf{Hence proved.}$$