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## unicwaan one year ago Can someone help me verify the identity? sine of x divided by one minus cosine of x + sine of x divided by one minus cosine of x = 2 csc x

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1. BTaylor

so, this? $\frac{\sin x}{1-\cos{x}} + \frac{\sin x}{1-\cos{x}} = 2\csc x$

2. unicwaan

yes except the it's 1- cos and 1 + cos sorry

3. welshfella

first convert the 2 fractions on left side to one fraction

4. BTaylor

You want to use the identity $$1-\cos^2 x = \sin^2 x$$. Multiply the 1 - cos x by (1 + cos x) on top and bottom, and you will have sin^2 x on the bottom.

5. unicwaan

Okay I understand that step BTaylpr but I don't know how the make the other fraction the same like what welshfella wants me to do

6. misty1212

HI!!

7. BTaylor

$\frac{\sin x}{1-\cos{x}} \cdot \frac{1+\cos x}{1+\cos x} + \frac{\sin x}{1+\cos{x}} \cdot \frac{1-\cos x}{1-\cos x} = 2\csc x$

8. misty1212

you can also just add, you will get it that way too

9. unicwaan

Oh the conjugate! I forgot about that!

10. misty1212

$\frac{b}{1+a}+\frac{b}{1-a}=\frac{b(1-a)+b(1+a)}{(1-a)(1+a)}$

11. misty1212

you get $\frac{2b}{1-a^2}$

12. unicwaan

Thank you I justed needed that step to get me started, I can figure the rest out :)

13. unicwaan

just*

14. BTaylor

no problem

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