I need help again

[Guest]

This is where I got stuck: The problem is once again verifying and proving one side equal to the other.

1-cos x/ 1+cos x= (csc x- cot x)^2

I turned the csc x into 1/sin x and the same with cot x. Then I tried to find the values of it squared and then I made them have a common denominator and I added and subtracted the values. I got no where. Please help me.

 

[Guest]

My time is running out!

I don't want to seem pushy or anything like that but my mom told me to turn off the computer at 8:55 so if you can please help me quickly. If not it's OK but please.
:!:

 

[Guest]

I must go !

I must go but please answer my question and I will look at it tommorow.

 

[soroban]

Hello, Carlos!

\(\displaystyle \L\frac{1\,-\,\cos x}{1\,+\,\cos x}\:=\\csc x\,-\,\cot x)^2\)

This is not an easy problem . . .
Here's one approach . . (the one you started)

The right side is: \(\displaystyle \,\left(\frac{1}{\sin x}\,-\,\frac{\cos x}{\sin x}\right)^2\;=\;\left(\frac{1\,-\,\cos x}{\sin x}\right)^2\;=\;\frac{(1\,-\,\cos x)^2}{\sin^2x}\)

Here's the magic step: \(\displaystyle \,\frac{(1\,-\,\cos x)^2}{1\,-\,cos^2x}\;=\;\frac{(1\,-\,cos x)^2}{(1\,-\,cos x)(1\,+\,\cos x)}\;=\;\frac{1\,-\,\cos x}{1\,+\,\cos x}\)