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