(c) $\dfrac{\sqrt{b^2 - a^2}}{b}$
Using $\sin^2\theta + \cos^2\theta = 1$: $\cos\theta = \sqrt{1 - \sin^2\theta} = \sqrt{1 - \dfrac{a^2}{b^2}} = \dfrac{\sqrt{b^2 - a^2}}{b}$
Source: Introduction to Trigonometry, Section 8.4
The key identity is $\sin^2\theta + \cos^2\theta = 1$. Substitute $\sin\theta = \frac{a}{b}$, square it, subtract from 1, and simplify. Examiners expect you to identify and apply the correct Pythagorean identity directly. Option (c) is the only one with the correct numerator $\sqrt{b^2-a^2}$ and denominator $b$.