Substituting standard values:
$$\cos 45° = \frac{1}{\sqrt{2}}, \quad \sin 60° = \frac{\sqrt{3}}{2}, \quad \sec 30° = \frac{2}{\sqrt{3}}, \quad \cosec 30° = 2$$
$$= \frac{\dfrac{1}{\sqrt{2}} + \dfrac{\sqrt{3}}{2}}{\dfrac{2}{\sqrt{3}} + 2} = \frac{\dfrac{\sqrt{2} + \sqrt{3}}{2\cdot\frac{1}{1}}{}}{}$$
Let me compute carefully:
Numerator: $\dfrac{1}{\sqrt{2}} + \dfrac{\sqrt{3}}{2} = \dfrac{2 + \sqrt{6}}{2\sqrt{2}}$
Denominator: $\dfrac{2}{\sqrt{3}} + 2 = \dfrac{2 + 2\sqrt{3}}{\sqrt{3}}$
$$= \frac{2+\sqrt{6}}{2\sqrt{2}} \times \frac{\sqrt{3}}{2+2\sqrt{3}} = \frac{(2+\sqrt{6})\sqrt{3}}{2\sqrt{2} \cdot 2(1+\sqrt{3})}$$
$$= \frac{2\sqrt{3}+\sqrt{18}}{4\sqrt{2}(1+\sqrt{3})} = \frac{2\sqrt{3}+3\sqrt{2}}{4\sqrt{2}(1+\sqrt{3})}$$
$$= \frac{\sqrt{2}(2\sqrt{3}+3\sqrt{2})}{4\sqrt{2}\cdot\sqrt{2}(1+\sqrt{3})\cdot\frac{\sqrt{2}}{\sqrt{2}}}{} $$
Multiply numerator and denominator by $\sqrt{2}$:
$$= \frac{\sqrt{2}(2\sqrt{3}+3\sqrt{2})}{8(1+\sqrt{3})} = \frac{2\sqrt{6}+6}{8(1+\sqrt{3})} = \frac{2(\sqrt{6}+3)}{8(1+\sqrt{3})} = \frac{\sqrt{6}+3}{4(1+\sqrt{3})}$$
$$= \frac{\sqrt{3}(\sqrt{2}+\sqrt{3})}{4(1+\sqrt{3})} \quad \Rightarrow \quad \boxed{\dfrac{\sqrt{3}(\sqrt{2}+\sqrt{3})}{4(1+\sqrt{3})}}$$
Rationalising: multiply by $\dfrac{({\sqrt{3}-1})}{(\sqrt{3}-1)}$... this simplifies to $\dfrac{3+\sqrt{3}}{4\cdot 2} = \dfrac{3+\sqrt{3}}{8}$
$$\therefore \quad \frac{\cos 45° + \sin 60°}{\sec 30° + \cosec 30°} = \dfrac{3+\sqrt{3}}{8}$$
Source: Introduction to Trigonometry, Section 8.3
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