Question

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

Answer 1

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

Answer 2

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

Answer 3

first convert the 2 fractions on left side to one fraction

Answer 4

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.

Answer 5

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

Answer 6

HI!!

Answer 7

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

Answer 8

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

Answer 9

Oh the conjugate! I forgot about that!

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

just*

Answer 14

no problem