Trig Identity Yay!

**Jkeeb** · February 13th 2011, 04:32 PM

Hey there. I have been trying to prove this trig identity is true for quite some time now with no success.

Here it is:

1-cosx/sinx + sinx/1-cosx = 2cscx

I am confused on how to elminate the fractions any insight would be appreciated thanks.

**skeeter** · February 13th 2011, 04:42 PM

Originally Posted by Jkeeb

Hey there. I have been trying to prove this trig identity is true for quite some time now with no success.

Here it is:

1-cosx/sinx + sinx/1-cosx = 2cscx

I am confused on how to elminate the fractions any insight would be appreciated thanks.

$\dfrac{1-\cos{x}}{\sin{x}} + \dfrac{\sin{x}}{1-\cos{x}} =<br />$

common denominator ...

$\dfrac{(1-\cos{x})(1-\cos{x}) + (\sin{x})(\sin{x})}{\sin{x}(1-\cos{x})} =<br />$

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

$\dfrac{2-2\cos{x}}{\sin{x}(1-\cos{x})} =$

$\dfrac{2(1-\cos{x})}{\sin{x}(1-\cos{x})} =$ ... you're there.

**Soroban** · February 13th 2011, 06:38 PM

Hello, Jkeeb!

Another approach . . .

$\dfrac{1-\cos x}{\sin x} + \dfrac{\sin x}{1-\cos x} \:=\: 2\csc x$

Multiply the second fraction by $\frac{1+\cos x}{1 + \cos x}$

. . $\displaystyle \frac{1-\cos x}{\sin x} + \frac{\sin x}{1-\cos x}\cdot\frac{1+\cos x}{1 + \cos x} \;=\; \frac{1-\cos x}{\sin x} + \frac{\sin x(1 + \cos x)}{1 - \cos^2\!x}$

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

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

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