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Mathematics 16 Online
OpenStudy (anonymous):

How do i verify 1+tan^3x / 1+tan^3x = 1-tanx+tan^2x

OpenStudy (anonymous):

can you rewrite that using the equation editor?

OpenStudy (abb0t):

Is this correct? \[\frac{ 1+\tan^3(x) }{ 1+\tan^3(x) }= 1-\tan(x)+\tan^2(x)\]

OpenStudy (anonymous):

that cant be right

OpenStudy (anonymous):

\[\frac{ 1+\tan^3x }{ 1+\tan^3x } = 1-tanx + \tan^2x\]

OpenStudy (anonymous):

o_o

OpenStudy (anonymous):

ahh i made a mistakeee!

OpenStudy (anonymous):

\[\frac{ 1+\tan^3x }{ 1+tanx } = 1-tanx + \tan^2\]

OpenStudy (anonymous):

oh ok ! that one looks like an identity . the last one did not. abb0t looks like he is about to help

OpenStudy (abb0t):

\[\tan^3(x) = \tan^2(x) \tan(x) = (\sec^2(x)-1)\tan(x)\]

OpenStudy (anonymous):

i think i got it. ill try to figure this out by myself!!

OpenStudy (anonymous):

okay...im really lost... ill draw it down what im doing so far

OpenStudy (anonymous):

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