I have to prove that:

[(1-cos x)/sin x] + [sin x/(1-cos x)] = 2 csc x

How would I go about doing this? I've tried several things, to no avail. Any help?

Thanks,

Jason

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- Oct 30th 2007, 08:43 AMJasonWProve identity problem
I have to prove that:

[(1-cos x)/sin x] + [sin x/(1-cos x)] = 2 csc x

How would I go about doing this? I've tried several things, to no avail. Any help?

Thanks,

Jason - Oct 30th 2007, 09:09 AMSoroban
Hello, Jason!

Quote:

Prove: .$\displaystyle \frac{1-\cos x}{\sin x} + \frac{\sin x}{1-\cos x}\; = \;2\csc x$

I've tried several things, to no avail.

__adding__those two fractions . . .

$\displaystyle \frac{1-\cos x}{\sin x} + \frac{\sin x}{1-\cos x} \;=\;{\color{blue}\frac{1-\cos x}{1-\cos x}}\cdot\frac{1-\cos x}{\sin x} + {\color{blue}\frac{\sin x}{\sin x}}\cdot\frac{\sin x}{1-\cos x}$

. . $\displaystyle = \;\frac{(1-\cos x)^2 + \sin^2x}{\sin x(1-\cos x)}\;=\;\frac{1 - 2\cos x + \overbrace{\cos^2x + \sin^2x}^{\text{This is 1}}}{\sin x(1 - \cos x)} $

. . $\displaystyle = \;\frac{2 - 2\cos x}{\sin x(1 - \cos x)} \;=\;\frac{2(1-\cos x)}{\sin x(1-\cos x)} \;=\;\frac{2}{\sin x} \;=\;2\csc x$

- Oct 30th 2007, 09:12 AMJasonW
Hehe, I didn't think it would be that simple :P

Thanks much!