find the area bounded by the curve C:y=tan x, tangent drawn to C at x=pi/4 and the x axis.
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What is the area of the region bounded by C: y = tanx, tangent drawn at C at x = pie/4 and the x-axis?
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Aryan Arora, Polymath/Sysadmin/Programmer/Mathematician/Philalethes etc.
Answered Nov 1, 2016
[math]x = \dfrac{\pi}{4}[/math]
[math] C : y = \tan \left(x\right) [/math]
[math]\implies y = \tan\left(\dfrac{\pi}{4}\right) = 1[/math]
tangent or slope drawn at any point at [math]C : m = \dfrac{dy}{dx} = \sec^2 \left(x\right)[/math]
[math]\implies m = \sec^2\left(\dfrac{\pi}{4}\right) = 2[/math]
now, the slope-intercept form of a line is [math]y=mx + b[/math] and we have the values of [math]y[/math], [math]m[/math]and [math]x[/math] , so plugging them in [math]y=mx + b[/math] will give us [math]b= 1-\dfrac{\pi}{2}[/math]
and, plugging just [math]b= 1-\dfrac{\pi}{2}[/math] and [math]m = 2[/math] in [math]y=mx + b[/math] will give us the tangent drawn at [math]x = \dfrac{\pi}{4}[/math] at [math]C[/math] , which is [math]y=2x + 1-\dfrac{\pi}{2}[/math] ,
now, graphing out [math]y=2x + 1-\dfrac{\pi}{2}[/math] , [math]y = \tan \left(x\right)[/math] and [math]x= \dfrac{\pi}{4}[/math]
( [math]x=\dfrac{\pi-2}{4}[/math] is the point [math]\text{M}[/math] because at [math]y=0[/math] , [math]y=2x + 1-\dfrac{\pi}{2} \implies x =\dfrac{\pi-2}{4} [/math]
and [math] x=\dfrac{\pi}{4}[/math] is the point [math]\text{N}[/math] ) :-
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3 ANSWERS

Aryan Arora, Polymath/Sysadmin/Programmer/Mathematician/Philalethes etc.
Answered Nov 1, 2016
[math]x = \dfrac{\pi}{4}[/math]
[math] C : y = \tan \left(x\right) [/math]
[math]\implies y = \tan\left(\dfrac{\pi}{4}\right) = 1[/math]
tangent or slope drawn at any point at [math]C : m = \dfrac{dy}{dx} = \sec^2 \left(x\right)[/math]
[math]\implies m = \sec^2\left(\dfrac{\pi}{4}\right) = 2[/math]
now, the slope-intercept form of a line is [math]y=mx + b[/math] and we have the values of [math]y[/math], [math]m[/math]and [math]x[/math] , so plugging them in [math]y=mx + b[/math] will give us [math]b= 1-\dfrac{\pi}{2}[/math]
and, plugging just [math]b= 1-\dfrac{\pi}{2}[/math] and [math]m = 2[/math] in [math]y=mx + b[/math] will give us the tangent drawn at [math]x = \dfrac{\pi}{4}[/math] at [math]C[/math] , which is [math]y=2x + 1-\dfrac{\pi}{2}[/math] ,
now, graphing out [math]y=2x + 1-\dfrac{\pi}{2}[/math] , [math]y = \tan \left(x\right)[/math] and [math]x= \dfrac{\pi}{4}[/math]
( [math]x=\dfrac{\pi-2}{4}[/math] is the point [math]\text{M}[/math] because at [math]y=0[/math] , [math]y=2x + 1-\dfrac{\pi}{2} \implies x =\dfrac{\pi-2}{4} [/math]
and [math] x=\dfrac{\pi}{4}[/math] is the point [math]\text{N}[/math] ) :-
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