# Tricky Trig Problem

1. Sep 3, 2010

### Char. Limit

1. The problem statement, all variables and given/known data
All right, so I was trying to help a friend prove a certain (complicated) trig identity for summer homework, but I got stuck myself... hopefully one of you will be able to help.

The trig identity in question is...

$$\frac{cos(x)}{1-tan(x)} + \frac{sin(x)}{1-cot(x)} = cos(x) + sin(x)$$

2. Relevant equations
1+tan^2(x)=sec^2(x)
1+cot^2(x)=csc^2(x)

3. The attempt at a solution

So far I've gotten it to...

$$\frac{cos(x)-sin(x)}{sec^2(x)-2tan(x)} - \frac{cos(x)-sin(x)}{csc^2(x)-2cot(x)} = cos(x)+sin(x)$$

But although I think that's a really nice form (two very similar terms), I have no idea where to go from there. Could one of you help me out?

2. Sep 3, 2010

### Dick

I would first express all of the trig functions in terms of sin(x) and cos(x) and then show the two sides are equal. It's pretty straightforward.

3. Sep 3, 2010

### eumyang

I would multiply the first fraction by
$$\frac{1 + \tan \,x}{1 + \tan \,x}$$
and multiply the second fraction by
$$\frac{1 + \cot \,x}{1 + \cot \,x}$$.

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