1. Given that $\sin \theta = \frac{a}{b}$, then cos θ is equal to : (a) $\frac{b}{\sqrt{b^2-a^2}}$ (b) $\frac{b}{a}$ (c) $\frac{\sqrt{b^2-a^2}}{b}$ (d) $\frac{a}{\sqrt{b^2-a^2}}$

Using $\sin^2\theta + \cos^2\theta = 1$ with $\sin \theta = \frac{a}{b}$ gives $\cos \theta = \frac{\sqrt{b^2 - a^2}}{b}$, which also matches the right-triangle picture ($\text{opposite} = a$, $\text{hypotenuse} = b$, $\text{adjacent} = \sqrt{b^2 - a^2}$).

Concept

• Fundamental identity: $\cos \theta = \sqrt{1 - \sin^2\theta}$ for acute $\theta$.
• Right-triangle view: $\sin \theta = \text{opposite}/\text{hypotenuse} = \frac{a}{b} \Rightarrow \text{adjacent} = \sqrt{b^2 - a^2}$ by Pythagoras.

Key steps

• $\cos \theta = \sqrt{1 - \frac{a^2}{b^2}} = \sqrt{\frac{b^2 - a^2}{b^2}} = \frac{\sqrt{b^2 - a^2}}{b}$ → option (c).

Common mistakes

• (a) $\frac{b}{\sqrt{b^2-a^2}}$: inverts numerator and denominator (looks like a secant-style flip).
• (b) $\frac{b}{a}$: this is cosec θ, the reciprocal of sin θ, not cos θ.
• (d) $\frac{a}{\sqrt{b^2-a^2}}$: this is tan θ (opposite/adjacent), not cos θ.

Real-world

• Safe method: always keep the largest side (hypotenuse, here b) in the denominator for both sin and cos; only tan places a side other than the hypotenuse in the denominator.