If sinθ = a/b then the value of secθ - cosθ is? (where 0° < θ < 90°)

Answer: Option C

$$\sin \theta =\frac{a}{b}=\frac{\text{P}}{\text{H}}$$

$$\begin{array}{rl}& \text{BC}=\sqrt{b^2-a^2}\\ & \left[\text{using pythagorad theorem}\right]\\ & \therefore sec\theta -\cos \theta \\ & =\frac{\text{H}}{\text{B}}-\frac{\text{B}}{\text{H}}\\ & =\frac{\text{AC}}{\text{BC}}-\frac{\text{BC}}{\text{AC}}\\ & =\frac{b}{\sqrt{b^2-a^2}}-\frac{\sqrt{b^2-a^2}}{b}\\ & =\frac{b^2-{\left(\sqrt{b^2-a^2}\right)}^2}{b\sqrt{b^2-a^2}}\\ & =\frac{b^2-b^2+a^2}{b\sqrt{b^2-a^2}}\\ & =\frac{a^2}{b\sqrt{b^2-a^2}}\end{array}$$