Question

The area bounded between the curve $y=\tan x$; tangent drawn to it at $x=\frac{\pi }{4}$ and $y\ge 0$ is

• A. $\frac{1}{4}({\log }_{e}4-1)$
• B. $\frac{1}{2}({\log }_{e}4-1)$
• C. $\frac{1}{2}({\log }_{e}4+1)$
• D. $\frac{1}{4}({\log }_{e}4+1)$

Answer 1

The correct option is A $\frac{1}{4}({\log }_{e}4-1)$

$y=\tan X$

now slope(m) of the line is ${\frac{dy}{dx}}_{(x=\frac{\pi }{4})}={\sec }^{2}X=2$

now in graph of $\tan X$ when $x=\frac{\pi }{4},y=1$

so equation of line is $(y-y_1)=m(x-x_1)$

so equation of line is $y=2x+1-\frac{\pi }{2}$

now at $y=0,x=\frac{\pi }{4}-\frac{1}{2}$

now $A={\int }_0^{\frac{\pi }{4}}\left(\tan X\right)-{\int }_{(\frac{\pi }{4}-\frac{1}{2})}^{\frac{\pi }{4}}(y=2x+1-\frac{\pi }{2})$

$A=\ln \left(\sec X\right)-[x^2-\frac{\pi x}{2}+x]$

now putting upper and lower limt,and on simplified the term, we will get

$A=\frac{(\ln 4-1)}{4}$