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LukecrayonzBest ResponseYou've already chosen the best response.0
cot x sec^4 x=cot x + 2 tan x + tan^3 x
 one year ago

LukecrayonzBest ResponseYou've already chosen the best response.0
Uhh what site is that.. :O
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
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
 one year ago

LukecrayonzBest ResponseYou've already chosen the best response.0
;__; you have different answers?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
there are different ways to do this, but they are both valid
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
notice how I'm only manipulating the right side and I'm changing it into the left side
 one year ago

LukecrayonzBest ResponseYou've already chosen the best response.0
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?
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
1/tan x and 2 tan^2 x/tan x are fractions with the same denominator (tan x)
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
I combined them to get (1 + 2 tan^2 x)/tan x
 one year ago

LukecrayonzBest ResponseYou've already chosen the best response.0
Thanks a lot @jim_thompson5910 :D
 one year ago

jim_thompson5910Best ResponseYou've already chosen the best response.1
you're welcome
 one year ago
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