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(D) Let $f(	heta) = a \sec 	heta - b 	an 	heta$.Then $f'(	heta) = a \sec 	heta 	an 	heta - b \sec^2 	heta = \sec 	heta (a 	an 	heta - b \s

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

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(D) Let $f(	heta) = a \sec 	heta - b 	an 	heta$.Then $f'(	heta) = a \sec 	heta 	an 	heta - b \sec^2 	heta = \sec 	heta (a 	an 	heta - b \sec 	heta)$.Setting $f'(	heta) = 0$,we get $a 	an 	heta - b \sec 	heta = 0$ (since $\sec 	heta 
eq 0$ for $0 This implies $a \sin 	heta = b$,or $\sin 	heta = \frac{b}{a}$.Since $\sin 	heta = \frac{b}{a}$,we have $\cos 	heta = \sqrt{1 - \frac{b^2}{a^2}} = \frac{\sqrt{a^2 - b^2}}{a}$.Thus,$\sec 	heta = \frac{a}{\sqrt{a^2 - b^2}}$ and $	an 	heta = \frac{b}{\sqrt{a^2 - b^2}}$.The second derivative $f''(	heta) = a \sec 	heta (1 + 2 	an^2 	heta) > 0$,confirming a minimum at $\sin 	heta = \frac{b}{a}$.The minimum value is $f(	heta) = a \left( \frac{a}{\sqrt{a^2 - b^2}} ight) - b \left( \frac{b}{\sqrt{a^2 - b^2}} ight) = \frac{a^2 - b^2}{\sqrt{a^2 - b^2}} = \sqrt{a^2 - b^2}$.

If $a$ and $b$ are positive numbers such that $a > b$,then the minimum value of $a \sec \theta - b \tan \theta$ for $0 < \theta < \frac{\pi}{2}$ is

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