What is the equation of the line tangent to # f(x)=-(sinx)/sin^2 x # at # x=1 #?

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Answer 1 · 2016-10-05T19:41:44

First note that #f(x)=-1/sinx = -cscx#

At #x=1#, we have #y = f(1) = -1/sin1 = -csc1#, so the tangent contains the point #(1,-csc1)#.

The slope of the tangent can be found by differentiating:

#f'(x) = cscxcotx#

At #x=1# the slope of the tangent line is #m=f'(1) = csc1 cot1#

The equation of the line through #(1, -csc1)# with slope #m = csc1 cot1# is

#y + csc1 = (csc1 cot1) (x - 1)#

In slope-intercept form, it is

#y = (csc1cot1) x-csc1cot1-csc1#