QJEEAdv08Ma7f0d: JEE Advanced 2008 Mathematics - Class 12 JEEAdv08Ma7f0d Solution: The area of the region between the curves and bounded by the…

Let us take the required equation

$y = \sqrt {{{1 + \sin x} \over {\cos x}}}$ and $y = \sqrt {{{1 - \sin x} \over {\cos x}}}$

$\int\limits_0^{{\pi \over 4}} {\left( {\sqrt {{{1 + \sin x} \over {\cos x}}} - \sqrt {{{1 - \sin x} \over {\cos x}}} } \right)dx}$

$\because$ \left( {{{1 + \sin x} \over {\cos x}} > {{1 - \sin x} \over {\cos x}} > 0} \right)

$= \int\limits_0^{{\pi \over 4}} {\left( {\sqrt {{{1 + {{2\tan {x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}}} \over {{{1 - {{\tan }^2}{x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}}}}} - \sqrt {{{1 + {{2\tan {x \over 2}} \over {1 + {{\tan }^2}{x \over 2}}}} \over {{{{{(1 - \tan {x \over 2})}^2}} \over {1 - \tan {x \over 2}}}}}} } \right)dx}$

$= \int\limits_0^{{\pi \over 4}} {\left( {\sqrt {{{{{\left( {1 + {{\tan }^2}{x \over 2}} \right)}^2}} \over {1 - {{\tan }^2}{x \over 2}}}} - \sqrt {{{{{\left( {1 - \tan {x \over 2}} \right)}^2}} \over {1 - {{\tan }^2}{x \over 2}}}} } \right)dx}$

$\int\limits_0^{{\pi \over 4}} {{{1 + \tan {x \over 2} - 1 + \tan {x \over 2}} \over {\sqrt {1 - {{\tan }^2}{x \over 2}} }}dx}$

$= \int\limits_0^{{\pi \over 4}} {{{2\tan {x \over 2}} \over {\sqrt {1 - {{\tan }^2}{x \over 2}} }}dx}$

Put $\tan {x \over 2} = t$

$dx = {{2\,dt} \over {1 + {t^2}}}$

$A = \int\limits_0^{\sqrt 2 - 1} {{{4t} \over {(1 + {t^2})\sqrt {1 - {t^2}} }}dt}$