verify the trigonometric equation by substituting identities to match the right hand side with the left hand side: (sinx / 1-cosx) + (sinx / 1+cosx) = 2cscx

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verify the trigonometric equation by substituting identities to match the right hand side with the left hand side: (sinx / 1-cosx) + (sinx / 1+cosx) = 2cscx

Mathematics
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looks like this \[\frac{ sinx }{ 1-cosx } + \frac{ sinx }{ 1+cosx }=2cscx\] been stuck on this forever...
on left hand side take L.C.M.
start with your left hand side :) hint: You'll want put the two fractions over the same denominator .

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so, would i get them both under 1-cos^2x?
over*
correct:)
It Becomes \[\frac{ sinx(1+cosx)+sinx(1-cosx) }{ \sin^2x }\]
alright then do I distribute sin?
yyeeepp
distribute sinx :)
\[\frac{ 2sinx }{ \sin^2x }\] like this?
yes
then from there, you can simplify the fraction and you'll get the answer (RHS) :)
Ok so how do I simplify this, Sorry! this is where i have been stuck at.
oh you can divide the numerator and denominator by sinx
so is it 2/sinx? And then just change it to csc?
likeee: |dw:1436251446732:dw|
and yes that's correct :)
ahh ok THANK YOUUU!!!!!!!!!!
no problem :)!!

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