Question

The area of the region enclosed between $y=tanx,\frac{\pi }{3}\le x\le \frac{\pi }{3},y=cotx,\frac{\pi }{6}\le x\le \frac{3\pi }{2}andX-$ axis is:

• A. $2log\sqrt{\frac{3}{2}}$
• B. $log\sqrt{\frac{3}{2}}$
• C. $2log\sqrt{\frac{2}{3}}$
• D. none of these

Answer 1

The correct option is A $2log\sqrt{\frac{3}{2}}$

$po\mathrm{int}of\mathrm{int}er\sec tionofgivencurves$
$\tan x=\cot x$
$\Rightarrow \tan x=\frac{1}{\tan x}$
$\Rightarrow {\tan }^{2}x=1$
$x=\pm {45}^0$
$rejecting-{45}^0\text{since it does not lies in the given region}$
$\text{Required Area=}\underset{\pi /6}{\overset{\pi /4}{\int }}\tan xdx+\underset{\pi /4}{\overset{\pi /3}{\int }}\cot xdx$
$={\left[\log \sec x\right]}_{\pi /6}^{\pi /4}+{\left[\log \sin x\right]}_{\pi /4}^{\pi /6}$
$=\log \sec \left(\frac{\pi }{4}\right)-\log \sec \left(\frac{\pi }{6}\right)+\log \sin \left(\frac{\pi }{6}\right)-\log \sin \left(\frac{\pi }{4}\right)$
$=\log \left(\sqrt{2}\right)-\log \left(\frac{2}{\sqrt{3}}\right)+\log \left(\frac{\sqrt{3}}{2}\right)-\log \left(\frac{1}{\sqrt{2}}\right)$
$=\log \left(\sqrt{2}\right)+\log \left(\frac{\sqrt{3}}{2}\right)+\log \left(\frac{\sqrt{3}}{2}\right)+\log \left(\sqrt{2}\right)[\log \left(\frac{1}{m}\right)=-\log m]$
$=2[\log \left(\sqrt{2}\right)+\log \left(\frac{\sqrt{3}}{2}\right)]$
$=2\left[\log \left(\frac{\sqrt{2}\sqrt{3}}{2}\right)\right][\log m+\log n=\log mn]$
$=2\left[\log \left(\sqrt{\frac{3}{2}}\right)\right]$