If sinθ = $$\frac{a}{b}$$  then the value of secθ - cosθ is? (where 0° < θ < 90°)

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

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