$\text{ The given curve is }y=\tan x$

$\therefore \frac{dy}{dx}={\sec }^{2}x$

${\left(\frac{dy}{dx}\right)}_{x=\pi /4}={\sec }^2\frac{\pi }{4}=2$

$\text{ Also, at }x=\frac{\pi }{4};y=1$

$\therefore \text{ The equation of the tangent to the curve at the point }$

$\left(\frac{\pi }{4},1\right)\text{ is }y-1=2\left(x-\frac{\pi }{4}\right)$

$y=0;x=\frac{\pi }{4}-\frac{1}{2}=OT$

$\text{ Now, the required area }=\text{ Area of curvilinear }\mathrm{△}OPN-\text{ Area of }\mathrm{△}PTN$

$={\int }_0^{\pi /4} (\tan x)dx-\frac{1}{2}\cdot NT\cdot PN$

$=[\log (\sec x){]}_0^{\pi /4}-\frac{1}{2}\left(\frac{\pi }{4}-\frac{\pi }{4}+\frac{1}{2}\right)\cdot 1=\frac{1}{2}\left(\log 2-\frac{1}{2}\right)$