Prove/show question

**Goldberg Variation** · October 5th 2009, 03:49 PM

Heya!

I've been struggling with this for some time now and would really appreciate some help. Need to show that the right side equals the left side.

$\frac{1}{sinx} = \frac{sinx}{1+cosx} + \frac{1}{tanx}$

Tried different ways to solve this but I get stuck each time.

RS = Rightside; sorry if this is not the right way to type this. Not familiar with the english terms

RS: $\frac{sinx}{1+cosx} + \frac{1}{\frac{sinx}{cosx}}$

After this I've experimented with changing 1's to $sin^2x+cos^2x$ among many other things.

Honestly feels like I am really on the wrong track here. Mostly the $1+cosx$ that annoys me.

Can someone shed some light on this for me? Thanks in advance!

**skeeter** · October 5th 2009, 04:35 PM

$\frac{\sin{x}}{1+\cos{x}} + \frac{\cos{x}}{\sin{x}}$

$\frac{\sin{x}}{1+\cos{x}} \cdot \frac{1-\cos{x}}{1-\cos{x}} + \frac{\cos{x}}{\sin{x}}<br />$

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

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

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

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

$\frac{1}{\sin{x}}$

**Goldberg Variation** · October 6th 2009, 01:04 AM

Thank you skeeter.

Didn't realize 1/tan = cotx = cosx/sinx. When knowing this it was pretty simple. Cheers!

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