Prove this guys,.,thanks 2 csc θ = 1 - cos θ + sin θ --------- -------- sin θ 1-cos θ
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is that supposed to be : $\displaystyle \frac{2csc\theta}{sin\theta}=\frac{1-cos\theta+sin\theta}{1-cos\theta}$ ?
Originally Posted by seld is that supposed to be : $\displaystyle \frac{2csc\theta}{sin\theta}=\frac{1-cos\theta+sin\theta}{1-cos\theta}$ ? But that is not an identity. Looking at the source it appears that: $\displaystyle 2\csc (\theta)= 1 -\frac{\cos(\theta)}{\sin(\theta)} +\frac{\sin(\theta)}{1-\cos(\theta)}$ is the indended expression, but that is also not an identity. CB
Last edited by CaptainBlack; Sep 15th 2009 at 01:40 AM. Reason: put back the 2 that got lost
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 . . .
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
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