## Lukecrayonz 2 years ago @jim_thompson5910 help with trig identities real quick?

1. Lukecrayonz

cot x sec^4 x=cot x + 2 tan x + tan^3 x

2. surdawi

sec^2=1+tan^2

3. Lukecrayonz

Steps?:O

4. zordoloom

Do you need help?

5. Lukecrayonz

Yes

6. zordoloom

That should help.

7. Lukecrayonz

Uhh what site is that.. :O

8. Lukecrayonz

@zordoloom

9. jim_thompson5910

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

10. Lukecrayonz

11. jim_thompson5910

there are different ways to do this, but they are both valid

12. jim_thompson5910

notice how I'm only manipulating the right side and I'm changing it into the left side

13. Lukecrayonz

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?

14. jim_thompson5910

1/tan x and 2 tan^2 x/tan x are fractions with the same denominator (tan x)

15. jim_thompson5910

I combined them to get (1 + 2 tan^2 x)/tan x

16. Lukecrayonz

Thanks a lot @jim_thompson5910 :D

17. jim_thompson5910

you're welcome