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Lukecrayonz
 4 years ago
@jim_thompson5910 help with trig identities real quick?
Lukecrayonz
 4 years ago
@jim_thompson5910 help with trig identities real quick?

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Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0cot x sec^4 x=cot x + 2 tan x + tan^3 x

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0Uhh what site is that.. :O

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1cot 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
 4 years ago
Best ResponseYou've already chosen the best response.0;__; you have different answers?

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1there are different ways to do this, but they are both valid

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1notice how I'm only manipulating the right side and I'm changing it into the left side

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.01/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
 4 years ago
Best ResponseYou've already chosen the best response.11/tan x and 2 tan^2 x/tan x are fractions with the same denominator (tan x)

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1I combined them to get (1 + 2 tan^2 x)/tan x

Lukecrayonz
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks a lot @jim_thompson5910 :D

jim_thompson5910
 4 years ago
Best ResponseYou've already chosen the best response.1you're welcome
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