Question

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

Answer 1

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

Answer 2

Finally a man here who knows how to use parentheses

Answer 3

what exactally are you looking for?

Answer 4

Verify the identity

Answer 5

lol my fault

Answer 6

Okay then, combine the fractions together first

Answer 7

2 sin(x) = 2 csc(x)

Answer 8

I know it's combing denominators first; it is just after that I need.

Answer 9

@OG_UnLeAsH: You forgot the denominator

Answer 10

darn it

Answer 11

So this identity is correct

Answer 12

okay @ayolisse can you draw out what you got after combining?

Answer 13

posted on wrong question whoops

Answer 14

sinx(1+cosx)+sinx(1-cosx)/(1-cosx)(1+cosx)

Answer 15

Distribute?

Answer 16

\[sinx+sinxcosx+sinx-sinxcosx/ (1-cosx)(1+cosx)\]

Answer 17

Add the like terms

Answer 18

\[2sinx+1-cosxsinx/ (1-cosx)(1+cosx)\]

Answer 19

Where did the 1 come from?

Answer 20

\[(\sin^2)x+(\cos^2)x=1\]

Answer 21

But there's no sin^2x or cos^2x

Answer 22

then what would you do cause now I am lost

Answer 23

Continue with \(\dfrac{\sin x+\sin x\cos x+\sin x-\sin x\cos x}{(1-\cos x)(1+\cos x)}\)

Answer 24

\[2sinx+cosxsinx-cosxsinx/(1-cosx)(1+cosx)\]

Answer 25

And then?

Answer 26

do the cosxsinx cancel?

Answer 27

Yes

Answer 28

now what about the denominator @kc_kennylau

Answer 29

Distribute

Answer 30

\[2sinx/1+cosx-cosx-(\cos^2)x\]

Answer 31

the cosx cancel right?

Answer 32

Yep

Answer 33

then what do you do after that?

Answer 34

Do you remember the identity you've said in the wrong time?

Answer 35

the (sin^2)x+(cos^2)x=1 ?

Answer 36

Yep

Answer 37

Make cos^2x the subject

Answer 38

do you make the denominator sin^2x?

Answer 39

Yes

Answer 40

okay i go the answer