- anonymous

sine of x divided by one minus cosine of x + sine of x divided by one minus cosine of x = 2 csc x

- chestercat

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- anonymous

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

- Nnesha

\[\huge\rm \frac{ \sin(x) }{ 1-cosine(x) } +\frac{ \sin(x) }{ 1 -cosine(x) }\]
what is the common denominator ?

- anonymous

1-cosine(x)?

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## More answers

- anonymous

but the second one is sinx/1+cosx

- Nnesha

well the common denominator is 1-cosine(x) obviously
both denominator are same u will get same numerator (sinx)\[\huge\rm \frac{ \sin(x) }{ 1-cosine(x) } \]

- Nnesha

huh uh-oh are you sure check your question again

- anonymous

yea i checked it 1+cosx

- Nnesha

http://prntscr.com/7vfzly

- Nnesha

\[\huge\rm \frac{ \sin(x) }{ 1-cosine(x) } +\frac{ \sin(x) }{ 1 +cosine(x) }\]
so common denominator is (1-cosinex)(1+cosinex)

- anonymous

but the picture of the equation in the book says 1+cosx

- Nnesha

alright you typed it wrong

- anonymous

it did that itself it copy pasted that way lol

- anonymous

##### 1 Attachment

- Nnesha

really ? :o you didin't type it ?

- Nnesha

shocking how that can translate :o :/

- anonymous

lol no that was weird tho right?

- Nnesha

\[\huge\rm \frac{ sinx(1+cosx) +sinx(1-cosx)}{ (1-cosx)(1+cosx)}\]
common denominator is (1-cosx)(1+cosx)
and multiply numerator of first fraction by denominator of 2nd fraction
*and multiply numerator of 2nd fraction by denominator of first fraction

- Nnesha

ye ....

- anonymous

and then you cancel out?

- Nnesha

no no NO!
no then u will get the original question

- Nnesha

distribute!

- anonymous

ohhh

- anonymous

wait what

- Nnesha

yes familiar withe the foil method ? apply that at the denominator

- anonymous

oh okay so i leave the numerator alone righ tnow

- anonymous

1-cos2x?

- Nnesha

yep right now distribute both parentheses by sinx 't the numerator

- anonymous

what? that last sentence you said threw me off i guess the wording did

- Nnesha

\[\huge\rm \frac{ \color{reD}{sinx}(1+cosx) +\color{red}{sinx}(1-cosx)}{ (1-cosx)^2}\]
distribute both parentheses by red sin x

- anonymous

1sinx+sinxcosx+1sinx-sinxcosx

- anonymous

2sinx/1-cos2x

- Nnesha

and cos^2x equal to what ?
do u remember sin/cos identities ?

- anonymous

csc?

- Nnesha

nope

- Nnesha

special identity \[\huge\rm sin^2x + \cos^2 x = 1\] solve this for cos^2x

- anonymous

some codes came out for when you said special identity nothing else?

- Nnesha

sin^2x + cos^2x = 1 solve for cos^2 x

- anonymous

cos^2x=1-sin^2x

- Nnesha

yes right so replace cos^2x by 1-sin^2x

- anonymous

okay

- Nnesha

\[\huge\rm \frac{ \color{reD}{2sinx}}{ 1-(1-sin^2x)}\]
distribute parentheses by negative one

- anonymous

its coding

- Nnesha

ahh
2sinx over 1-(1-sin^2x)

- anonymous

sorry lol im being a pain

- anonymous

2sinx/sin^2x

- Nnesha

yes sin^2x is same as sinx times sin x
so 2sin over sinx sin x
simplfy

- anonymous

1/sin2x

- Nnesha

is it sin^2 x ?

- anonymous

yea but with one over it

- Nnesha

no what about 2 ?
it's 2sinx over sinx times sin x

- Nnesha

|dw:1437522158545:dw|

- anonymous

it all cancels out to 1/sinx

- Nnesha

no what about 2 ??|dw:1437522218631:dw|

- anonymous

im so lost i dont like this number two lol

- Nnesha

just look at that drawing who kidnapped that 2 ?

- anonymous

the sinx? i have no idea

- Nnesha

sin x cancels out right ??
so where is 2 ?
look at there

- anonymous

2 over sinx

- Nnesha

ahh there ugo!

- Nnesha

i was about ...to .. put u in jail
how....darrrrrrrrrrrrrre u kidnapped 2 :/
;)
hey gtg gO_OD luck!

- anonymous

haha okay but what do i do to finish?

- Nnesha

well 2 over sinx is same as 2 times 1/sinx
and 1/sinx = csc
2 times csc

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