trig

tjd:

(1-(sin^2)x)/(cot^2)x=

hi tdk
I think that you have a bracket in a funny spot.
Before do any thing else I am going to show you how to use the sup button that is located above the smilies.
sup stands for superscript.

Write sin then hit the sup button. the curser will go exactly where it wants you to enter 2 then just click to the end of the function and keep going.
so your question is

( 1- sin ²x ) / cot ²x

What is this bit 1- sin ²x ?
Maybe it has got some thing to do with
sin ²x + cos ²x = 1 ?

What about cot x.
isn't cot x = 1/tanx ?
And, when you divide isn't that the same as multiplying by the reciprocal?

That's enough hints.
You see if you can do it.

So for this we can use trig identities.
The one I will be using is:

1 = sin^2(x) + cos^2(x)

because of this we can simplify this equation to:

(cos^2(x))/(cot^2(x)) =

from there we can break up cotangent into:

cot^2(x) = (cos^2(x))/(sin^2(x))

allowing us to have the equation simplify to:

(cos^2(x))/[(cos^2(x))/(sin^2(x))] ---> which you should be able to do yourself