Question

sin x /1 - cos x + sin x / 1- cos x = 2 csc x

Answer 1

\[\frac{2sin(x)}{1-cos(x)}=\frac{2}{sin(x)}\]

Answer 2

with me?

Answer 3

No how did you get 2 sin x

Answer 4

divided by 1 -cos x

Answer 5

You still here ?

Answer 6

Its proving trig equations if that helps solve it

Answer 7

sorry had to run to the store
\[\frac{sin(x)}{1-cos(x)}+\frac{sin(x)}{1-cos(x)}\]

Answer 8

is this what you have?

Answer 9

@Seadog12 ?

Answer 10

yeah but the denominator of the last one is 1 + cosx

Answer 11

ahhh
\[\frac{sinx}{1-cos(x)}+\frac{sinx}{1+cosx}\]
find common denominator
\[\frac{sin(x)(1+cos(x))+sin(x)(1-cos(x))}{(1-cos(x))(1+cos(x))}=\\\frac{sin(x)+sin(x)cos(x)+sin(x)-cos(x)sin(x)}{(1-cos^2(x))}=\frac{2sin(x)}{sin^2(x)}=\frac{2}{sin(x)}=\\2csc(x)\]

Answer 12

understand?

Answer 13

so the cos cancel out on the top

Answer 14

the cosines cancel out

Answer 15

well look at the second to last line
sin(x)+sin(x)cos(x)+sin(x)−cos(x)sin(x)
rearrange
sin(x)cos(x) - sin(x)cos(x) + sin(x) + sin(x)
(sin(x)cos(x)-sin(x)cos(x)) + 2sin(x)
= ( 0 ) + 2sin(x)
= 2sin(x)

Answer 16

Ok

Answer 17

Thanks zzr0ck3r