Question

The area of the region between the curves
$y=\sqrt{\frac{1+sinx}{cosx}}$ and $y=\sqrt{\frac{1-sinx}{cosx}}$ and bounded by the
lines $x=0$ and $x=\frac{\pi }{4}$ is

• A. $(a)\underset{0}{\overset{\sqrt{2}-1}{\int }}\frac{t}{(1+t^2)\sqrt{1-t^2}}dt$
• B. $(b)\underset{0}{\overset{\sqrt{2}-1}{\int }}\frac{4t}{(1+t^2)\sqrt{1-t^2}}dt$
• C. $(c)\underset{0}{\overset{\sqrt{2}+1}{\int }}\frac{4t}{(1+t^2)\sqrt{1-t^2}}dt$
• D. $(d)\underset{0}{\overset{\sqrt{2}+1}{\int }}\frac{t}{(1+t^2)\sqrt{1-t^2}}dt$

Answer 1

Answer: (B)

Required area $=\underset{0}{\overset{\pi /4}{\int }}(\sqrt{\frac{1+sinx}{cosx}}-\sqrt{\frac{1-sinx}{cosx}})dx$
$[\because \frac{1+sinx}{cosx}>\frac{1-sinx}{cosx}>0]$
$=\underset{0}{\overset{\pi /4}{\int }}(\sqrt{\frac{1+\frac{2tan\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}}{\frac{1-ta{n}^2\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}}}-\sqrt{\frac{1-\frac{2tan\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}}{\frac{1-ta{n}^2\frac{x}{2}}{1+ta{n}^2\frac{x}{2}}}})dx$
$=\underset{0}{\overset{\pi /4}{\int }}(\sqrt{\frac{1+tan\frac{x}{2}}{1-tan\frac{x}{2}}}-\sqrt{\frac{1-tan\frac{x}{2}}{1+tan\frac{x}{2}}})dx$
$=\underset{0}{\overset{\pi /4}{\int }}\frac{1+tan\frac{x}{2}-1+tan\frac{x}{2}}{\sqrt{1-ta{n}^2\frac{x}{2}}}dx=\underset{0}{\overset{\pi /4}{\int }}\frac{2tan\frac{x}{2}}{\sqrt{1-ta{n}^2\frac{x}{2}}}dx$
Put $tan\frac{x}{2}=t\Rightarrow \frac{1}{2}se{c}^2\frac{x}{2}dx=dt$
$=\underset{0}{\overset{tan\pi /8}{\int }}\frac{4tdt}{(1+t^2)\sqrt{1-t^2}}$
As $\underset{0}{\overset{\sqrt{2}-1}{\int }}\frac{4tdt}{(1+t^2)\sqrt{1-t^2}}[\because tan\frac{\pi }{8}=\sqrt{2}-1]$