One complicated trig identity, Someone help! (tan x) / (sin x- cos x) = (sin^2 x + (sin x)(cos x))/ (cos x - 2 cos ^3 x)

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One complicated trig identity, Someone help! (tan x) / (sin x- cos x) = (sin^2 x + (sin x)(cos x))/ (cos x - 2 cos ^3 x)

Mathematics
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LHS: sin --- cos ------- sin - cos ---- 1 flip the bottom over and multiply to get...
sin --------- = RHS cos(sin-cos) RHS: (sin^2 x + (sin x)(cos x))/ (cos x - 2 cos ^3 x) sin(sin + cos) ------------- cos(1 - 2cos^2)
1 sin + cos tan ------- = tan -------- (sin - cos) 1 - 2cos^2 divide both sides by tan:......

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Other answers:

multiply both side by (-1) to get: -1 sin + cos ------- = --------- sin-cos 2cos^2 -1
cross multiply....im sure :) 2cos^2 -1 = 2(sin+cos)
2 cos^2 -sin^2 -cos^2 = 2(sin + cos) cos ^2 - sin^2 = 2(sin+cos)....
sin+cos(sin - cos) = sin^2 - cos^2 NOT (2)sin+cos
go back to the cross multiply...
Thank you for showing me, but unfortunately I'm required to work on one side
if you can work it out to meet in the middle, then you simply reverse it to get back to the one side.... How you solve it really doesnt matter a hilll of beans :)
ok, thanks :)
but let me show you that it DOES work out :) I gots: -(2cos^2 -1) = sin^2 - cos^2 -(2cos^2 -(sin^2 + cos^2)) = sin^2 - cos^2 -(2cos^2 -sin^2 -cos^2) = sin^2 - cos^2 -(cos^2 - sin^2) = sin^2 - cos^2 sin^2 - cos^2 = sin^2 - cos^2 (proofed)
I'm kinda confused now...
who requires you ta work only one side?
my professor requires us to work one side to equal other side
since both sides are equally "messed up"; it would be best to "simplify" both sides till they are equal. I you can only work with one die, you are pretty much tieing your hands behind your back.....
my keyboard hates me :)
ok, its fine lol.. i appreciate the help, it would be easier to work both sides.. wish i could

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