If $$a > b > 0,$$   the minimum value of $$a\sec \,\theta - b\tan \,\theta $$    is :

Solution :
$$\eqalign{ & f\left( \theta \right) = a\sec \,\theta - b\tan \,\theta \cr & \therefore f'\left( \theta \right) = a\sec \,\theta .\tan \,\theta - b{\sec ^2}\theta \cr & \therefore f'\left( \theta \right) = 0 \Rightarrow \sec \,\theta \left( {a\tan \,\theta - b\sec \,\theta } \right) = 0 \cr & \Rightarrow \sec \,\theta = 0{\text{ or }}a\tan \,\theta = b\sec \,\theta \cr & {\text{But }}\sec \,\theta \ne 0 \cr & \therefore f'\left( \theta \right) = 0 \cr & \Rightarrow a\tan \,\theta = b\sec \,\theta \,\,\,\,\,\,\,\, \Rightarrow b = \frac{{a\tan \,\theta }}{{\sec \,\theta }} = a\sin \,\theta \cr & \therefore \sin \,\theta = \frac{b}{a} \cr & f''\left( \theta \right) = a\sec \,\theta .{\tan ^2}\theta + a{\sec ^3}\theta - 2b{\sec ^2}\theta .\tan \,\theta \cr & = a.\frac{a}{{\sqrt {{a^2} - {b^2}} }}.\frac{{{b^2}}}{{{{\left( {\sqrt {{a^2} - {b^2}} } \right)}^2}}} + a.{\left( {\frac{a}{{\sqrt {{a^2} - {b^2}} }}} \right)^3} - 2b{\left( {\frac{a}{{\sqrt {{a^2} - {b^2}} }}} \right)^2}.\frac{b}{{\sqrt {{a^2} - {b^2}} }} \cr & = \frac{1}{{{{\left( {{a^2} - {b^2}} \right)}^{\frac{3}{2}}}}}.\left\{ {{a^2}{b^2} + {a^4} - 2{a^2}{b^2}} \right\} \cr & = \frac{{{a^2}\left( {{a^2} - {b^2}} \right)}}{{{{\left( {{a^2} - {b^2}} \right)}^{\frac{3}{2}}}}} \cr & = \frac{{{a^2}}}{{\sqrt {{a^2} - {b^2}} }} > 0 \cr & \therefore \,\min f\left( \theta \right) = a.\frac{a}{{\sqrt {{a^2} - {b^2}} }} - b.\frac{b}{{\sqrt {{a^2} - {b^2}} }} \cr & = \frac{{{a^2} - {b^2}}}{{\sqrt {{a^2} - {b^2}} }} \cr & = \sqrt {{a^2} - {b^2}} \cr} $$