What is the equation of the line normal to # f(x)=secx-cot^2x# at # x=pi/3#?

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Answer 1 · 2018-05-12T12:37:03

The equation of normal is #3 x +15 y = 28.14#

#sec (pi/3)= 2 , tan (pi/3)= sqrt3 , cot (pi/3)= 1/sqrt3#

# cot^2 (pi/3)= 1/3 ,csc (pi/3)= 2/sqrt3, csc^2 (pi/3)= 4/3,#

#f(x) = sec x- cot ^2 x :. f(pi/4) = sec (pi/4)- cot ^2 (pi/4)# or

# f(pi/4) = 2-1/3 =5/3# The point at which normal to be drawn

is #(pi/3 , 5/3) ; f(x)=sec x- cot ^2 x #

#f'(x)=sec x tan x - 2 cot x * (-csc^2x) # or

#f'(x)=sec x tan x + 2 cot x * csc^2 x #

#f'(pi/4)=sec (pi/4) tan (pi/4) + 2 cot (pi/4) * csc^2 (pi/4) #or

#f'(pi/4)=2sqrt3 + 2/sqrt3 * 4/3 = 5.00 # The slope of

tangent at #pi/3# is #5.0# therefore, slope of normal at #pi/3#

is #(-1/5)# The equation of normal to #f(x)# at point

#(pi/3 , 5/3)# is # y- 5/3 = -1/5 (x- pi/3)# or

#15 y- 25 = -3 x + pi or 3 x +15 y = 25 +pi# or

#3 x +15 y = 28.14#

The equation of normal is #3 x +15 y = 28.14# [Ans]