Find the tangent to curve y=2cot(3x)-3cosec(4x) when x =pi/6

I’m going back and fourth with myself in my because my working out seems far too long….

So, the gradient I got is

F’(x) = 12cosec(4x)cot(4x)-6cosec^2(3x)

At x=pi/6

F’(pi/6)=12cosec(2pi/3)cot(2pi/3)-6cosec^2(pi/2)


The y coordinate for the tangent is:

2cot(pi/2)-3cosec(2pi/3) when x =pi/6


The answer asks for in exact form….

Am I meant to leave it such form as above? Or would I simply y to root form : -2root3 as 2cot(pi/2) is invalid

Sorry, I mean undefined. What’s the difference? Just so I know in layman’s terms lol.


Yeah, I finally got there in the end y=-14x+7pi/3-2root3

I had trouble with this one but I’ve been full of flu, so I assume it’s because of that