@jim_thompson5910 help with trig identities real quick?

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cot x sec^4 x=cot x + 2 tan x + tan^3 x
sec^2=1+tan^2
Steps?:O

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

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