Question

PLEASE HELP!!!!!!!!

Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation.

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

Answer 1

@kaiidayy What is this?

Answer 2

@Butterfly16 : \[\frac{\sin x}{1-\cos x} + \frac{\sin x}{1+\cos x}\]

Answer 3

Like this?

Answer 4

Yes

Answer 5

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

Answer 6

\[\frac{sinx +\sin x \cos x + \sin x - \cos x \sin x}{1 - \cos^2 x}\]

Answer 7

\[\frac{2 \sin x}{ 1 - \cos^2 x }\]

Answer 8

Can you solve now?

Answer 9

Do they turn into a different identity?

Answer 10

\[\frac{ 2 \sin x }{ 1-\cos^2 x }\]is equal to 2 csc x right?

Answer 11

Recall
cos^2 x + sin^2 x =1

Answer 12

I don't understand how to put that into the equation :/

Answer 13

cos^2 x + sin^2 x =1

sin^2x = 1 - cos^2 x

Answer 14

so:

2sinx/sin^2x

Answer 15

And what would that go down to? I cant find an identity in my notes that has those..

Answer 16

this is the basic identity.

Answer 17

Would I use sin(x)=1/csc(x)?

Answer 18

\[\Large \frac{2sinx}{\sin^2x } \to \frac{2 \sin x}{\sin x *\sin x}\]

Answer 19

cscx = 1/sinx
Yes.

Answer 20

so the sin x would cancel one on the top and give sin x?

Answer 21

you'll be left behind with:
2/sin x

Answer 22

and that would be 2 csc x right? since 1/sin x is csc then 2/sin x would make it 2 csc x?

Answer 23

Yes sir.

Answer 24

Thank you SO much for your help! I really truly appreciate it. :o)

Answer 25

:)