# Math Help - Trig Identity Yay!

1. ## Trig Identity Yay!

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.

2. 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}} =
$

common denominator ...

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

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

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

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

3. Hello, Jkeeb!

Another approach . . .

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

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

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

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

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