### (a)

**Step 1 — Write the standard values.**
$$\cos 60^\circ = \frac12,\quad \sec 30^\circ = \frac{2}{\sqrt3},\quad \tan 45^\circ = 1$$

**Step 2 — Simplify the denominator.**
$$\sin^2 30^\circ + \cos^2 30^\circ = 1 \qquad (\text{identity } \sin^2\theta+\cos^2\theta=1)$$

**Step 3 — Simplify the numerator.**
$$5\left(\frac12\right)^2 + 4\left(\frac{2}{\sqrt3}\right)^2 - (1)^2 = \frac54 + 4 \times \frac43 - 1 = \frac54 + \frac{16}{3} - 1$$

**Step 4 — Take LCM = 12.**
$$= \frac{15 + 64 - 12}{12} = \frac{67}{12}$$

**Step 5 — Divide by the denominator.**
$$\frac{67/12}{1} = \frac{67}{12}$$

$$\boxed{\dfrac{67}{12}}$$

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### (b)

**Step 1 — Take the LHS and replace $\cot^2\alpha$ using the identity $\operatorname{cosec}^2\alpha - \cot^2\alpha = 1$.**
$$\text{LHS} = 1 + \frac{\operatorname{cosec}^2\alpha - 1}{1 + \operatorname{cosec}\alpha}$$

**Step 2 — Factorise the numerator as a difference of squares.**
$$= 1 + \frac{(\operatorname{cosec}\alpha - 1)(\operatorname{cosec}\alpha + 1)}{1 + \operatorname{cosec}\alpha}$$

**Step 3 — Cancel $(1 + \operatorname{cosec}\alpha)$.**
$$= 1 + (\operatorname{cosec}\alpha - 1)$$

**Step 4 — Simplify.**
$$= \operatorname{cosec}\alpha = \text{RHS}$$

$$\textbf{Hence proved.}$$