The area of the region (in square units) above the x-axis bounded by the curve y= tanx,0≤ x≤ π 2 and the tangent to the curve at x= π 4 is

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

Also, $\frac{dy}{dx}=se{c}^{2}x$dxdy​=sec2x and ${(\frac{dy}{dx})}_{x=\frac{\pi }{4}}=se{c}^{2}(\frac{\pi }{4})={(\sqrt{2})}^{2}=2$(dxdy​)x=4π​​=sec2(4π​)=(2​)2=2

We know that the equation of the tangent to a curve $y=f(x)$y=f(x) at a point $({x}_{1},{y}_{1})$(x1​,y1​) is $y-{y}_{1}={(\frac{dy}{dx})}_{x=\frac{\pi }{4}}(x-{x}_{1})$y−y1​=(dxdy​)x=4π​​(x−x1​)

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

$$\Rightarrow y-1=2x-\frac{\pi }{2}$$⇒y−1=2x−2π​

$$\Rightarrow y-1+\frac{\pi }{2}=2x$$⇒y−1+2π​=2x

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

$$2x=\frac{\pi }{2}-1$$2x=2π​−1

$$\Rightarrow x=\frac{\pi }{4}-\frac{1}{2}$$⇒x=4π​−21​

Thus, the point $A\equiv (\frac{\pi }{4}-\frac{1}{2},0).$A≡(4π​−21​,0).

Now, the graph for the given information is

And, the required Area

$$={\int }_{0}^{\frac{\pi }{4}}(\mathrm{tan}x)dx-ar(\Delta PAB)$$=∫04π​​(tanx)dx−ar(ΔPAB)

$$={[\mathrm{log}(\mathrm{sec}x)]}_{0}^{\frac{\pi }{4}}-\frac{1}{2}\times (PB\times AB)$$=[log(secx)]04π​​−21​×(PB×AB)

$$=[\mathrm{log}(\mathrm{sec}\frac{\pi }{4})-\mathrm{log}(\mathrm{sec}0)]-\frac{1}{2}\times 1\times (\frac{\pi }{4}-(\frac{\pi }{4}-\frac{1}{2}))$$=[log(sec4π​)−log(sec0)]−21​×1×(4π​−(4π​−21​))

$$=|\mathrm{log}(\sqrt{2})-\mathrm{log}(1)|-\frac{1}{4}$$=∣log(2​)−log(1)∣−41​

$$=|\mathrm{log}{(2)}^{\frac{1}{2}}-0|-\frac{1}{4}$$=∣log(2)21​−0∣−41​

$=\frac{1}{2}\mathrm{log}2-\frac{1}{4}=\frac{1}{2}(\mathrm{log}2-\frac{1}{2})$=21​log2−41​=21​(log2−21​) sq units.