What is the equation of the line tangent to #f(x)=-tan(x - pi/2) # at #x=pi/3#?

1 Answer
Ratnaker Mehta
Jul 7, 2016

#12x+9y-3sqrt3-4pi=0.#

Explanation:

#f(x)=-tan(x-pi/2) =-{-tan(pi/2-x)}=tan(pi/2-x)=cotx rArr f(pi/3)=cot(pi/3)=1/sqrt3#

Now, #f(x)=cotx rArr f'(x)=-cosec^2x rArr f'(pi/3)=-cosec^2(pi/3)=-4/3.#

But, #f'(x)#= slope of tgt. at pt. #(x,f(x))#

#:.# Slope tgt. at pt. #(pi/3,f(pi/3))=(pi/3,1/sqrt3)# is #-4/3,# and, tgt. passes thro. pt. #(pi/3,1/sqrt3)#.

Since we know slope and a pt. on tgt., we can obtain its eqn. using slope-pt. form.

#:.# Reqd. eqn. of tgt. is #: y-1/sqrt3=-4/3(x-pi/3),# i.e., #9y-3sqrt3+4(3x-pi),# or, #12x+9y-3sqrt3-4pi=0.#

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