1. ## Prove this

Prove this guys,.,thanks

2 csc θ = 1 - cos θ + sin θ
--------- --------
sin θ 1-cos θ

2. is that supposed to be :

$\frac{2csc\theta}{sin\theta}=\frac{1-cos\theta+sin\theta}{1-cos\theta}$ ?

3. Originally Posted by seld
is that supposed to be :

$\frac{2csc\theta}{sin\theta}=\frac{1-cos\theta+sin\theta}{1-cos\theta}$ ?
But that is not an identity.

Looking at the source it appears that:

$2\csc (\theta)= 1 -\frac{\cos(\theta)}{\sin(\theta)} +\frac{\sin(\theta)}{1-\cos(\theta)}$

is the indended expression, but that is also not an identity.

CB

4. Originally Posted by CaptainBlack
But that is not an identity.

CB

true, but that's all I could make of the original post with the

----- -------
sin x 1-cosx

doesn't make any sense to me to have (sin x)*1-cos x as a denominator.

looking at the source? err . . . the 2 dissapeared . . .

5. Originally Posted by seld
true, but that's all I could make of the original post with the

----- -------
sin x 1-cosx

doesn't make any sense to me to have (sin x)*1-cos x as a denominator.

looking at the source? err . . . the 2 dissapeared . . .
If you had quoted what the OP had posted you will have seen what they typed, and been better able to interpret it.

I have put the lost 2 back where it belongs now.

CB