# Precalculus HW problem

1. Oct 25, 2006

### pasatom20

Can anyone help me sove this problem?

cot x + tan x + 1 = (cot x / 1 - tan x) + (tan x / 1 - cot x)

Last edited: Oct 25, 2006
2. Oct 25, 2006

convert the right hand sides to sines and cosines, add the fractions, and simplify. or you could use the fact that tan x = 1 / cot x

Last edited: Oct 25, 2006
3. Oct 25, 2006

### lkh1986

Since right hand side is more complex, I think we should start from that side, then arrive at the left hand side.

We can use the properties of that cot x is the reciprocal of tan x, too.

4. Oct 25, 2006

### pasatom20

Here is what i got by working on the right hand side.

(cot x / 1- (sin x / cos x)) + (tan x / 1 - (cos x / sinx)) = (cot x / (cos x - sin x / cos x)) + (tan x / (sin x - cos x / sin x))

then i tried several different ways to work on the problem from this step, but it never worked.

5. Oct 25, 2006

convert the numerators to sines and cosines as well.

6. Oct 25, 2006

### pasatom20

here is what i got

(cos x / sin x ) / (cos x - sin x / cos x ) + (sin x / cos x) / (sin x - cos x / sin x)
= (cos x / sin x )*(cos x / cos x - sin x) + (sin x / cos x)*(sin x / sin x - cos x)
= (cos ^2 x / cos x - sin^2 x) + (sin^2 x / sin x - cos^2 x)
= (sin^2 x - 1 / cos x - sin^2 x ) + (cos^2 x - 1 / sin x - cos^2 x)
I tried to work on the problem until i got to this step, but I'm not sure if I'm on the right track.

7. Oct 25, 2006

So $$\cot x + \tan x + 1 = \frac{\cot x}{1- \frac{1}{\cot x}}$$ etc..