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\[Tan^2x - \cos^2x = 1/\cos^2 - 1 - \cos^2x\]

can you rewrite that right side? is that \[\frac 1{cos^2 x - 1} - \cos ^2 x\]

|dw:1352207602501:dw|

you know this identity right? \[\sin^2 x = 1 - \cos^2 x\]

yes

wait..nevermind what i said....

alright

yes :) I was on the right track whew :P

now combine the numerators \[\frac{1 - \cos^2 x - \cos ^ 4 x}{\cos^2 x}\]
is this what you did?

no. that works too \[\frac{1 - \cos^2 x}{\cos^2 x} - \cos^2 x\]
yes?

hmm, why didnt you just totally cancel out the denominator?

I was left with 1-cos^2x-cos^2x :S

ah. that's different

you can't do that

|dw:1352208030577:dw|

hmm whats the reason I couldnt?

|dw:1352208100756:dw|

ap calc etcher tell me that

OH I understand

yes! I love you :P

should i move to the left side?

wo think i got it? L.S) will equel 1-cos^2/cos^2x - Cos^2x

tan^3 x = tan^2 x * tan x

that can work

or maybe tan^3 x = sin^3 x / cos^3 x

Thanks guys, lemme try this other one and i cant get it ill post a new question

all the best :)