The area of the region ( in square units ) above the x -axis bounded by the curve y = tan x ,   0…

The given curve is $y=\tan x$ and at $x=\frac{\mathrm{\pi }}{4}, y=\tan \left(\frac{\mathrm{\pi }}{4}\right)=1$

Also, $\frac{dy}{dx}=se{c}^{2}x$ and ${\left(\frac{dy}{dx}\right)}_{x=\frac{\mathrm{\pi }}{4}}=se{c}^2\left(\frac{\mathrm{\pi }}{4}\right)={\left(\sqrt{2}\right)}^2=2$

We know that the equation of the tangent to a curve $y=f\left(x\right)$ at a point $\left(x_1, y_1\right)$ is $y-y_1={\left(\frac{dy}{dx}\right)}_{x=\frac{\mathrm{\pi }}{4}}\left(x-x_1\right)$

Hence, the equation of the tangent to $y=\tan x$ at $P\left(\frac{\mathrm{\pi }}{4}, 1\right)$ is $y-1=2\left(x-\frac{\mathrm{\pi }}{4}\right)$

$\Rightarrow y-1=2x-\frac{\mathrm{\pi }}{2}$

$\Rightarrow y-1+\frac{\mathrm{\pi }}{2}=2x$

To find the point where this line cuts the $x$-axis, put $y=0,$ to get

$2x=\frac{\mathrm{\pi }}{2}-1$

$\Rightarrow x=\frac{\mathrm{\pi }}{4}-\frac{1}{2}$

Thus, the point $A\equiv \left(\frac{\mathrm{\pi }}{4}-\frac{1}{2}, 0\right).$

Now, the graph for the given information is

And, the required Area

$={\int }_0^{\frac{\mathrm{\pi }}{4}}\left(\tan x\right)\mathrm{d}x-ar\left(∆PAB\right)$

$={\left[\log \left(\sec x\right)\right]}_0^{\frac{\mathrm{\pi }}{4}}-\frac{1}{2}\times \left(PB\times AB\right)$

$=\left[\log \left(\sec \frac{\mathrm{\pi }}{4}\right)-\log \left(\sec 0\right)\right]-\frac{1}{2}\times 1\times \left(\frac{\mathrm{\pi }}{4}-\left(\frac{\mathrm{\pi }}{4}-\frac{1}{2}\right)\right)$

$=\left|\log \left(\sqrt{2}\right)-\log \left(1\right)\right|-\frac{1}{4}$

$=\left|\log {\left(2\right)}^{\frac{1}{2}}-0\right|-\frac{1}{4}$

$=\frac{1}{2}\log 2-\frac{1}{4}=\frac{1}{2}\left(\log 2-\frac{1}{2}\right)$ sq units.