The area of the region between the curves \(y=\sqrt{\frac{1+\sin x}{\cos x}}\) and \(y=\sqrt{\frac{1-\sin x}{\cos x}}\) bounded by the lines \(x=0\) and \(x=\frac{\pi}{4}\) is

\[
\begin{aligned}
& \text { Required area }=\int_0^{\pi / 4}\left(\sqrt{\frac{1+\sin x}{\cos x}}-\sqrt{\frac{1-\sin x}{\cos x}}\right) d x \quad\left[\because \frac{1+\sin x}{\cos x}>\frac{1-\sin x}{\cos x}>0\right] \\
& \left.=\int_0^{\pi / 4} \sqrt{\frac{1+\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}{\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}} \sqrt{\frac{1-\frac{2 \tan \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}{\frac{1-\tan ^2 \frac{x}{2}}{1+\tan ^2 \frac{x}{2}}}}\right) d x \\
& \left.=\int_0^{\pi / 4} \sqrt{\frac{1+\tan \frac{x}{2}}{1-\tan \frac{x}{2}}}-\sqrt{\frac{1-\tan \frac{x}{2}}{1+\tan \frac{x}{2}}}\right) d x \\
&
\end{aligned}
\]

\[
=\int_0^{\pi / 4} \frac{1+\tan \frac{x}{2}-1+\tan \frac{x}{2}}{\sqrt{1-\tan ^2 \frac{x}{2}}} d x=\int_0^{\pi / 4} \frac{2 \tan \frac{x}{2}}{\sqrt{1-\tan ^2 \frac{x}{2}}} d x
\]
Put \(\tan \frac{x}{2}=t\)
\[
\begin{aligned}
\Rightarrow \quad \frac{1}{2} \sec ^2 \frac{x}{2} d x & =d t \\
\therefore \text { Required area } & =\int_0^{\tan \frac{\pi}{8}} \frac{4 t d t}{\left(1+t^2\right) \sqrt{1-t^2}} \\
& =\int_0^{\sqrt{2}-1} \frac{4 t}{\left(1+t^2\right) \sqrt{1-t^2}} d t \quad\left[\text { astan } \frac{\pi}{8}=\sqrt{2}-1\right]
\end{aligned}
\]