Question

can someone help me with this
cotxsec^4(x)=cotx+2tanx+tan^3(x)

Answer 1

Let's start with the left side.

Answer 2

can you express the left side in terms of sin and cos?

Answer 3

alternatively, if you want to do it without converting everything to sin and cos,
you can just start by writing
\(\sec^4 x = (\sec^2 x)^2\)
using the pythagorean identity for sec^2 and then expanding :)

this method could be faster :)

Answer 4

I'll do both methods, even though I have to agree with @hartnn
(It's been a while since I played with these identity problems)

Answer 5

The identity is \[1+\tan^2=\sec^2\]

Answer 6

I'll skip writing x for my own convenience

Answer 7

hold on, ive been working on it let me show you what i got

Answer 8

Sure.

Answer 9

1+tan^3(x)=cos^2(x)+2sinx/2cosx+tan^3(x)

Answer 10

How did you get to there?
you can just say it simply

Answer 11

how

Answer 12

\[\cot(\sec^2)^2=\cot(1+\tan^2)^2=\cot(1+2\tan^2+\tan^4)=\cot+2\tan+\tan^3\]

Answer 13

That's the left side

Answer 14

which is equal to the right side
the proof is complete

Answer 15

?

Answer 16

im confused

Answer 17

in that proof,
we used the fact that
since, tan x = 1/cot x

cot x * tan x =1

Answer 18

oh sorry
forgot to explain that

Answer 19

4 steps are shown as the proof...which step are you confused at ?

Answer 20

are you guys going based on the original problem or the simplified one i put?

Answer 21

the original problem

Answer 22

notice that \[\cot*\sec^4=\cot(\sec^2)^2\]

Answer 23

but no where in the original problem does it say cotx*tanx
it says cotx+2tanx

Answer 24

from third to fourth step...\[\cot(1+2\tan^2+\tan^4)=\cot*1+\cot*2\tan^2+\cot*\tan^4=\cot+2\tan*\tan*\cot+\]\[\cot*\tan*\tan^3=\cot+2\tan+\tan^3\]

Answer 25

do you have any questions?

Answer 26

nope let me just go over it right quick

Answer 27

sure. take as much time as needed

Answer 28

im so sorry but im just not getting it

Answer 29

I'm really sorry, but I have to go.

Answer 30

no problem thanks though