What is the equation of the line tangent to # f(x)=(x-cosx)/(x-secx) # at # x=pi/3#?

Question

What is the equation of the line tangent to #  f(x)=(x-cosx)/(x-secx)  # at #  x=pi/3#?

Isabella Ackerman · Accepted Answer

#6(pi-6)^2y-3(2pi-3)(pi-6)=(15sqrt3pi-27-36sqrt3)(3x-pi)#

The slope of the tangent is the same as the slope of curve at that point, which is given by the value of derivative of the function at that point.

Hence, slope of tangent to curve #f(x)=(x-cosx)/(x-secx)# at #x=pi/3# is given by #f'(pi/3)#. And as the tangent passes through #(x,f(x))# i.e. #(pi/3,f(pi/3))#, the equation of tangent is

#y-f(pi/3)=f'(pi/3)(x-pi/3)#

Now #f(pi/3)=(pi/3-cos(pi/3))/(pi/3-sec(pi/3))=(pi/3-1/2)/(pi/3-2)=(2pi-3)/(2pi-12)#

and using quotient rule #f'(x)=((x-secx)(1+sinx)-(x-cosx)(1-secxtanx))/(x-secx)^2#

= #(x+xsinx-secx-tanx-x+xsecxtanx+cosx-tanx)/(x-secx)^2#

= #(xsinx-secx-2tanx+xsecxtanx+cosx)/(x-secx)^2#

and #f'(pi/3)=(pi/3xxsqrt3/2-2-2sqrt3+pi/3xx2xxsqrt3+1/2)/(pi/3-2)^2#

= #(3sqrt3pi-36-36sqrt3+12sqrt3pi+9)/(2(pi-6)^2)#

= #(15sqrt3pi-27-36sqrt3)/(2(pi-6)^2)#

and equation of tangent is

#y-(2pi-3)/(2pi-12)=(15sqrt3pi-27-36sqrt3)/(2(pi-6)^2)(x-pi/3)#

or #(2pi-12)y-(2pi-3)=(15sqrt3pi-27-36sqrt3)/(pi-6)(x-pi/3)#

or #6(pi-6)^2y-3(2pi-3)(pi-6)=(15sqrt3pi-27-36sqrt3)(3x-pi)#

graph{(6(pi-6)^2y-3(2pi-3)(pi-6)-(15sqrt3pi-27-36sqrt3)(3x-pi))(y-(x-cosx)/(x-secx))=0 [-5, 5, -2.5, 2.5]}

William Abdalla · Answer

undefined The equation of the line tangent to f(x) = (x - cosx)/(x - secx) at x = pi/3 is y = -2x + 2√3 + 2π/3.

Isaiah Allen · Answer

undefined To find the equation of the line tangent to $ f(x) = \frac{x - \cos x}{x - \sec x} $ at $ x = \frac{\pi}{3} $, we need to find the derivative of $ f(x) $ and then evaluate it at $ x = \frac{\pi}{3} $ to find the slope of the tangent line. Then, we use the point-slope form of the equation of a line to find the equation of the tangent line.

$ f(x) = \frac{x - \cos x}{x - \sec x} $

$ f'(x) = \frac{(1 + \sin x)(x - \sec x) - (1 - \cos x)(1 - x \sec x \tan x)}{(x - \sec x)^2} $

Evaluating $ f'(x) $ at $ x = \frac{\pi}{3} $:

$ f'\left(\frac{\pi}{3}\right) = \frac{(1 + \sin \frac{\pi}{3})\left(\frac{\pi}{3} - \sec \frac{\pi}{3}\right) - (1 - \cos \frac{\pi}{3})\left(1 - \frac{\pi}{3} \sec \frac{\pi}{3} \tan \frac{\pi}{3}\right)}{\left(\frac{\pi}{3} - \sec \frac{\pi}{3}\right)^2} $

Now, we can calculate the value of $ f'\left(\frac{\pi}{3}\right) $. This will give us the slope of the tangent line.

After finding the slope, we can use the point-slope form of the equation of a line with the point $ \left(\frac{\pi}{3}, f\left(\frac{\pi}{3}\right)\right) $ and the slope we just calculated to find the equation of the tangent line.