NIMCET 2021 Question Paper with solutions

Given that : There are two curves
$\mathit{y}=\tan \mathit{x}\left(\frac{-\mathrm{\pi }}{3}\le \mathit{x}\le \frac{\mathrm{\pi }}{3}\right)\mathrm{}$
$\mathit{y}=\cot \mathit{x}\left(\frac{\mathrm{\pi }}{6}\le \mathit{x}\le \frac{3\mathrm{\pi }}{2}\right)$
To Find : The area of region bounded by the $\mathit{x}$-axis.
Point of intersection of $\mathit{y}=\tan \mathit{x}$ and $\mathit{y}=\cot \mathit{x}$
$\therefore \tan \mathit{x}=\cot \mathit{x}$
$\therefore \mathit{x}=\pm \frac{\mathrm{\pi }}{4}$
Since,$\mathit{x}=-\frac{\mathrm{\pi }}{4}\in \left(\frac{\mathrm{\pi }}{6},\frac{3\mathrm{\pi }}{2}\right)$
Let, $\mathit{x}=\frac{\mathrm{\pi }}{4}$ be the intersection point.
Now, we will find area of shaded region.
Area $=\underset{\frac{\mathrm{\pi }}{6}}{\overset{\frac{\mathrm{\pi }}{4}}{\int }}\tan \mathit{x}\mathit{d}\mathit{x}+\underset{\frac{\mathrm{\pi }}{4}}{\overset{\frac{\mathrm{\pi }}{3}}{\int }}\cot \mathit{x}\mathit{d}\mathit{x}$
$={\left[\log \sec \mathit{x}\right]}_{\mathrm{\pi }∕4}^{\mathrm{\pi }∕4}+{\left[\log \sin \mathit{x}\right]}_{\mathrm{\pi }∕4}^{\mathit{\pi }∕3}$
$=\left[\log \sec \frac{\mathrm{\pi }}{4}-\log \sec \frac{\mathrm{\pi }}{6}\right]$ $+\left[\log \sin \frac{\mathrm{\pi }}{3}-\log \sin \frac{\mathrm{\pi }}{4}\right]$
$=\left[\log \sqrt{2}-\log \frac{2}{\sqrt{3}}\right]$ $+\left[\log \frac{\sqrt{3}}{2}-\log \frac{1}{\sqrt{2}}\right]$
$=\log \sqrt{2}+\log \frac{\sqrt{3}}{2}+\log \frac{\sqrt{3}}{2}+\log \sqrt{2}$
$=\log \sqrt{2}+2\log \frac{\sqrt{3}}{2}$
$=\log 2+\log \frac{3}{4}$
$=\log \left(2\times \frac{3}{4}\right)$
$=\log \frac{3}{2}$ squnit
Hence, the required area is $=\log \frac{3}{2}$ squnit.