If \(\mathrm{a}\) and \(\mathrm{b}\) are positive number such that \(\mathrm{a}>\mathrm{b}\), then the minimum value of \(a \sec \theta-b \tan \theta\left(0 < \theta < \frac{\pi}{2}\right)\) is

\(\text {let } \mathrm{f}(\theta)=\mathrm{a} \sec \theta-b \tan \theta \)
\( \therefore \mathrm{f}^{\prime}(\theta)=\mathrm{a} \sec \theta \tan \theta-b \sec ^2 \theta \)
\(=\sec \theta(\mathrm{a} \tan \theta-\mathrm{b} \sec \theta) \)
\( \therefore \mathrm{f}^{\prime}(\theta)=0 \Rightarrow \sec \theta(\mathrm{a} \tan \theta-\mathrm{b} \sec \theta)=0 \)
\( \Rightarrow \mathrm{a} \tan \theta-\mathrm{b} \sec \theta=0 \quad \ldots\left[\text { As } 0 < \pi < \frac{\pi}{2}, \sec \theta \neq 0\right] \)
\( \Rightarrow \mathrm{a} \sin \theta-b=0 \quad \ldots\left[\text { As } 0 < \pi < \frac{\pi}{2}, \cos \theta \neq 0\right]\)
\(\Rightarrow \sin \theta=\frac{b}{a} \)
\(\Rightarrow \sec \theta=\frac{a}{\sqrt{a^2-b^2}} \text { and } \tan \theta=\frac{b}{\sqrt{a^2-b^2}}\)
Now,
\(\mathrm{f}^{\prime \prime}(\theta)=\sec \theta \tan \theta(\mathrm{a} \tan \theta-\mathrm{b} \sec \theta) \) \(+\sec \theta\left(a \sec ^2 \theta-b \sec \theta \tan \theta\right) \)
\(=\mathrm{a} \tan ^2 \theta \sec \theta-\mathrm{b} \sec ^2 \theta \tan \theta \) \(+a \sec ^3 \theta-b \sec ^2 \theta \tan \theta \)
\(=a \sec \theta\left(\tan ^2 \theta+\sec ^2 \theta\right) \)
\(=a \sec \theta\left(1+2 \tan ^2 \theta\right) \)
\(>0 \quad \ldots\left[\because \text { a is positive and } 0 < \theta < \frac{\pi}{2}\right] \)
\(f(\theta)\) is minimum when \(\sin \theta=\frac{b}{a}\).
\(\therefore \ \text {Minimum value of } f(\theta) \)
\( =a\left(\frac{a}{\sqrt{a^2-b^2}}\right)-b\left(\frac{b}{\sqrt{a^2-b^2}}\right) \quad \ldots[\text { From (i) }] \)
\( =\frac{a^2-b^2}{\sqrt{a^2-b^2}} \)
\( =\sqrt{a^2-b^2}\)