Finding the Equation of a Tangent

[The Matador]

Homework Statement

Find the equation of the tangent at the point indicated
y=3cscx
x=pi/4

2. The attempt at a solution

So to do the question you need a point (which I am given), the slope and then you need to substitute that all into y=mx+b.

I believe I have differentiated correctly with

y=3(-cot(x)csc(x))

Then I find the slope by inserting pi/4 into the differentiated equation

y=3(-cot(pi/4)csc(pi/4))

After this is done I put everything into y=mx+b, and this is where my problem is. Is there anyways to simplify 3csc(pi/4) and 3(-cot(pi/4)csc(pi/4))? I think I remember doing something similar in math a long time ago but I forgot now. The final answer does not include -cot or csc so there must be a way I don't remember.

 

[Gib Z]

What is cosec and cotangent in terms of the other trig functions?

1/(sin x) = csc x
1/( tan x) = cot x

So 3 csc (pi/4) = \frac{3}{\sin (\pi/4)}.

The exact value for sin (pi/4) is 1/(sqrt2) so..
3 \csc (\pi/4) = \frac{3}{\sqrt{2}} = \frac{3\sqrt{2}}{2}

So the same for cot, knowing tan (pi/4) = 1.