Lukecrayonz
@jim_thompson5910 help with trig identities real quick?
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Lukecrayonz
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cot x sec^4 x=cot x + 2 tan x + tan^3 x
surdawi
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sec^2=1+tan^2
Lukecrayonz
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Steps?:O
zordoloom
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Do you need help?
Lukecrayonz
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Yes
zordoloom
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That should help.
Lukecrayonz
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Uhh what site is that.. :O
Lukecrayonz
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@zordoloom
jim_thompson5910
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cot x sec^4 x=cot x + 2 tan x + tan^3 x
cot x sec^4 x=1/tan x + 2 tan x + tan^3 x
cot x sec^4 x=1/tan x + 2 tan^2 x/tan x + tan^3 x
cot x sec^4 x=(1 + 2 tan^2 x)/tan x + tan^3 x
cot x sec^4 x=(1 + 2 tan^2 x)/tan x + tan^3 x/1
cot x sec^4 x=(1 + 2 tan^2 x)/tan x + tan^4 x/tan x
cot x sec^4 x=(1 + 2 tan^2 x + tan^4 x)/tan x
cot x sec^4 x=(1 + 2z + z^2)/tan x ... Let z = tan^2 x
cot x sec^4 x=((z+1)^2)/tan x
cot x sec^4 x=((tan^2 x+1)^2)/tan x
cot x sec^4 x=((sec^2)^2)/tan x
cot x sec^4 x=(sec^4 x)/tan x
cot x sec^4 x=cot x sec^4 x
Lukecrayonz
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;__; you have different answers?
jim_thompson5910
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there are different ways to do this, but they are both valid
jim_thompson5910
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notice how I'm only manipulating the right side and I'm changing it into the left side
Lukecrayonz
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1/tan x + 2 tan^2 x/tan x + tan^3 x
(1 + 2 tan^2 x)/tan x + tan^3 x
What allowed you to do that?
jim_thompson5910
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1/tan x and 2 tan^2 x/tan x are fractions with the same denominator (tan x)
jim_thompson5910
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I combined them to get (1 + 2 tan^2 x)/tan x
Lukecrayonz
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Thanks a lot @jim_thompson5910 :D
jim_thompson5910
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you're welcome