Help with a trig proof...

**bemidjibasser** · June 18th 2009, 09:03 AM

I would appreciate some help proving the following:

(sin of theta/1-cos theta) - (1+cos theta/ sin theta)= 0
Thanks for looking, and thanks in advance.

**great_math** · June 18th 2009, 09:18 AM

(sin of theta/1-cos theta) - (1+cos theta/ sin theta)= 0

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

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

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

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

$=0=R.H.S$

**SENTINEL4** · June 18th 2009, 09:26 AM

You have

$\frac{sin\vartheta}{1-cos\vartheta}-\frac{1+cos\vartheta}{sin\vartheta}$

You multiply the first fraction with $sin\vartheta$ and the second with $1-cos\vartheta$ so you get...

$\frac{sin^2\vartheta}{(1-cos\vartheta)sin\vartheta}-\frac{(1+cos\vartheta)(1-cos\vartheta)}{(1-cos\vartheta)sin\vartheta}$ = $\frac{sin^2\vartheta}{(1-cos\vartheta)sin\vartheta}-\frac{1-cos\vartheta+cos\vartheta-cos^2\vartheta}{(1-cos\vartheta)sin\vartheta}$ = $\frac{sin^2\vartheta+cos^2\vartheta-1}{(1-cos\vartheta)sin\vartheta}$

But we know that $sin^2\vartheta+cos^2\vartheta=1$ so

$\frac{1-1}{(1-cos\vartheta)sin\vartheta}$ = $\frac{0}{(1-cos\vartheta)sin\vartheta} = 0$

**bemidjibasser** · June 18th 2009, 10:04 AM

thank you both for the quick and useful help!

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Help with a trig proof...

thanks!

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