Question

Find the area of the region bounded by the curve y=tan x,tangent drawn to C at x=$\pi /4$ and the x-axis.

Answer 1

The curve $C:y=tanx⟶\left(i\right)$

At $x=\frac{\pi }{4},y=1$ i.e $(\frac{x}{4},1)$ is a point on $C$.

From $\left(1\right)$, $\frac{dy}{dx}={\sec }^{2}x$, At $x=\frac{\pi }{4},\frac{dy}{dx}={\sec }^2\frac{\pi }{4}=2$.

The equation of the tangent to the curve $\left(1\right)$ at $x=\frac{\pi }{4}$ is

$y-1={\left[\frac{dy}{dx}\right]}_{x=\pi /4}(x-\frac{\pi }{4})$ or, $y-1=2(x-\frac{\pi }{4})$ or, $y=2x+1\frac{\pi }{2}⟶\left(2\right)$

If the tangent meet the $x$ axis at $O$, then for $O,y=0$,

$\therefore$ $2x+1-\frac{\pi }{2}=0$, or $x=\frac{\pi }{4}-\frac{1}{2}$ $\therefore$ $OO=\frac{\pi }{4}-\frac{1}{2}$.

Since $OR=\frac{\pi }{4}$ ; $\therefore$ $OR=OR-OO=\frac{\pi }{4}-(\frac{\pi }{4}-\frac{1}{2})=\frac{1}{2}$,

Also $PR=1$

The required area is shaded in figure

Hence the required are.

$=$ area of the region $OPQ$

$=$ (area of the region $OPR$) $-$ (area of $\Delta PQR$)

$={\int }_0^{\pi /4}ydx-\frac{1}{2}QR-PR$

$={\int }_0^{\pi /4}tanxdx\frac{-1}{2},\frac{1}{2},1={\left[logsecx\right]}_0^{\pi /4}-\frac{1}{4}$

$=(log\sqrt{2}-log1)-\frac{1}{4}=\frac{1}{2}log2-\frac{1}{4}=\frac{1}{2}(log2-\frac{1}{2})$ Sq units.