# Thread: Another simplify trig. problem.

1. ## Another simplify trig. problem.

Okay, now I'm new to this forum so should I make a topic for separate questions or just bump up my old one with a new post?

Anywho, it's rather simple but I want to verify if the way I did the process was correct. Er, I don't know how to post those.. image things so i'll just give you the problem and if it's possible, can you show me the process of doing it?

The problem is simplifying the following:

cot^2 - 4 / cot ^ 2 - cot -6

I know I can simplify cot into cos/sin but after simplifying I'm confused as to what to do with the fractions. I find a common denominator which is sin for both the top and the bottom, then I multiply by the bottom part of the fraction but my answer just seems a little wonky. Thanks in advance.

2. You're thinking this question is harder than it is. All you need to do is factorise and cancel. As follows:

$\displaystyle \frac{\cot^2 x -4}{\cot^2 x - \cot x - 6}$
$\displaystyle = \frac{(\cot x - 2)(\cot x +2)}{(\cot x -3)(\cot x + 2)}$
$\displaystyle = \frac{\cot x - 2}{\cot x -3}$

That should be all that's required.

3. ..That's it? Damn it, I need to stop over thinking these things x]. Thanks.