Explanation

(b)
$Required\ area = \int_{0}^{\pi/4}\left( \sqrt{\frac{1 + \sin x}{\cos x}} - \sqrt{\frac{1 - \sin x}{\cos x}} \right)dx$
$\because\left\lbrack \ \frac{1 + \sin x}{\cos x} > \frac{1 - \sin x}{\cos x} > 0 \right\rbrack$
$= \int_{0}^{\pi/4}\left( \sqrt{\frac{1 + \frac{2\tan\frac{x}{2}}{1 + \tan^{2}\frac{x}{2}}}{\frac{1 - \tan^{2}\frac{x}{2}}{1 + \tan^{2}\frac{x}{2}}} -}\sqrt{\frac{1 - \frac{2\tan\frac{x}{2}}{1 + \tan^{2}\frac{x}{2}}}{\frac{1 - \tan^{2}\frac{x}{2}}{1 + \tan^{2}\frac{x}{2}}}} \right)dx$
$= \int_{0}^{\pi/4}{\frac{1 + \tan{\frac{x}{2} - 1 + \tan\frac{x}{2}}}{\sqrt{1 - {tan\ }^{2}\frac{x}{2}}}dx =}\int_{0}^{\pi/4}{\frac{2\tan\frac{x}{2}}{\sqrt{1 - {tan\ }^{2}\frac{x}{2}}}dx}$
$put\ \tan{\frac{x}{2} = t \Longrightarrow \frac{1}{2}\sec^{2}}\frac{x}{2}dx = dt$
$\therefore Required\ area = \int_{0}^{\tan\frac{\pi}{8}}\frac{4t\ \ dt}{\left( 1 + t^{2} \right)\sqrt{1 - t^{2}}}$
$= \int_{0}^{\sqrt{2} - 1}{\frac{4t\ \ }{\left( 1 + t^{2} \right)\sqrt{1 - t^{2}}}dt}$
$\left\lbrack \because\tan{\frac{\pi}{8} = \sqrt{2} - 1} \right\rbrack$