If sin $$\theta=\frac{a}{b}$$, then the value of sec $$\theta-cos\ \theta$$ is (where $$0^\circ<\theta<90^\circ$$)

Given : sin $$\theta=\frac{a}{b}$$

Thus, using $$cos\theta=\sqrt{1-sin^2\theta}$$

=> $$cos\theta=\sqrt{1-(\frac{a}{b})^2}$$

=> $$cos\theta=\frac{\sqrt{b^2-a^2}}{b}$$

To find : sec $$\theta-cos\ \theta$$

= $$\frac{1}{cos\theta}-cos\theta$$

= $$\frac{1-cos^2\theta}{cos\theta}=\frac{sin^2\theta}{cos\theta}$$

= $$(\frac{a^2}{b^2})\div(\frac{\sqrt{b^2-a^2}}{b})$$

= $$(\frac{a^2}{b^2})\times(\frac{b}{\sqrt{b^2-a^2}})$$

= $$\frac{a^{2}}{b\sqrt{b^{2}-a^{2}}}$$

=> Ans - (D)