Question

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

Answer 1

These just keep getting better and better :D
\[\Large \frac{\sin x}{1-\cos x}+\frac{\sin x}{1+\cos x}=2\csc x\]

Answer 2

yes that is the equation. Please help me.

Answer 3

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

Answer 4

ok.

Answer 5

Well, don't be a stranger ^.^
simplify it...
Distribute what's to be distributed :D

Answer 6

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

Answer 7

Wrong. The second fraction, why is the sin x negative? :/

Answer 8

ohhhh so the second one looks like:

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

Answer 9

bbfoster123, keep in mind that (a-b)(1+b) = a^2-b^2
and that 1-cos^2 = sin^2

Answer 10

So...
\[\frac{ \sin x+ \cos x\sin x }{ (1-\cos x)(1+\cos x) } + \frac{\sin x- \cos x\sin x }{(1-\cos x)(1+\cos x) }\]
They now have the same denominator, you can simply combine the numerators :)
Do this now :D

Answer 11

thanks @jdoe0001

@PeterPan the equation would now be:

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

Answer 12

Yes... now, you can simplify the denominator as @jdoe0001 instructed
(a - b)(a + b) = a² - b²

And you can also simplify the numerator further~ by combining like terms ^.^
Do this now...

Answer 13

so it is:\[\frac{ \sin^2x + \cos^2x \sin^2x }{ \sin^2 x }\]

Answer 14

No... try again :)
We're adding the numerators, not multiplying :P

Answer 15

\[\frac{ 2\sin x }{ \sin^2x }\]

Answer 16

Much better :)
Can anything be cancelled out? ^.^

Answer 17

sin x = 1/csc x

Answer 18

Yeah, sure, but if you have
\[\frac{2a}{a^2}\]doesn't one of the a's cancel out?

Answer 19

yes they do. so it is sin x/ sin^2 x

Answer 20

So...
\[\huge \frac{2\sin x}{\sin^2 x}\]becomes...?

Answer 21

\[\frac{ sinx }{ \sin^2 x }\]

Answer 22

Why'd you remove the 2?

Answer 23

arent we simplifying?

Answer 24

Yes, we are. So how did the 2 disappear?

Answer 25

\[\frac{ 2\sin x }{ \sin x }\]

Answer 26

Now, you forgot that the sin at the bottom was squared.

Answer 27

\[\frac{ 2\sin x }{ \sin ^2 x }\]

Answer 28

Now... cancel out what CAN be cancelled out :)

Answer 29

so its \[\frac{ \sin x }{ \sin x } = \frac{ 1/\csc }{ 1/\csc } \]

Answer 30

then you mutiply and you get 2csc x

Answer 31

If it works for you, then :)

Answer 32

it doesn't work at all.

Answer 33

hello...

Answer 34

Because... as I said, you only have to cancel out. Don't convert it to csc yet.\[\frac{ 2\sin x }{ \sin ^2 x }\]

Answer 35

the cancelling part confuses me.

Answer 36

How would you simplify
\[\large \frac{2a}{a^2}\]?

Answer 37

\[\frac{ 2 }{ a }\]

Answer 38

Well, I don't really see what's so hard about that, so how to simplify
\[\Large \frac{ 2\sin x }{ \sin ^2 x }\]?

Answer 39

\[\frac{ 2 }{ \sin x }\]

Answer 40

See? Simple ^.^
Now what's 1/sin x?

Answer 41

csc

Answer 42

And there you have it :)

Answer 43

Thank You!!!!!!