What is the equation of the line that is normal to #f(x)=cscx+cotx # at # x=-pi/3#?

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Answer 1 · 2016-07-26T11:05:19

#3x-6y+pi-6sqrt3=0#.

We know that #dy/dx=f'(x)# gives the slope of tgt. to curve # C : y=f(x)# at any pt.#P(x,y)# on the curve.

since, normal line is #bot# to tgt., its slope, at any pt. #P(x,y)#, will be #-1/(f'(x)), if f'(x)!=0#

The Curve is # C : y=f(x)=cscx+cotx#

#:. f'(x)=-cscxcotx-csc^2x=-cscx(cotx+cscx)#

#:. f'(-pi/3)=-csc(-pi/3){cot(-pi/3)+csc(-pi/3)}#

#=-2/sqrt3(1/sqrt3+2/sqrt3)=-6/3=-2#

Therefore, the slope of normal #=1/2#.

Also, #f(-pi/3)=csc(-pi/3)+cot(-pi/3)=-sqrt3#.

Thus, the normal line passes thro. pt. #(-pi/3,-sqrt3)# and has slope#=1/2#.

Therefore, the eqn. of normal is given by,

#y+sqrt3=1/2(x+pi/3)#, i.e., #6y+6sqrt3=3x+pi#, or

#3x-6y+pi-6sqrt3=0#.