Question

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

Answer 1

I got as far as the right side =\[\frac{\cos ^{4}(x)+2\sin ^{2}(x)\cos ^{2}(x)+\sin ^{4}(x) }{ \sin(x)\cos ^{3}(x) }\]
But after that I'm confused on what to do

Answer 2

Start with the LHS

Answer 3

\(\cot x \sec^4x = \cot x (\sec^2x)(\sec^2x)\)

Right?

Answer 4

no, the right hand side is all addition, not multiplication

Answer 5

We will get there

Answer 6

wait

Answer 7

i get it sorry

Answer 8

You mean you can figure it out from here?

Answer 9

no im not sure on that

Answer 10

i just get what you wrote

Answer 11

\(\sec^2x = \tan^2 + 1\) correct?

Answer 12

yes!

Answer 13

So

\(\cot x(\sec^2x)(\sec^2x) = \cot x(\tan^2x + 1)(\tan^2x +1)\) Do you agree?

Answer 14

yeah, but where do you go from there?

Answer 15

do you have to work on the other side or stay o the LHS?

Answer 16

If we expand \((\tan^2x + 1)(\tan^2x + 1)\) what do you get from that?

Answer 17

No, LHS only

Answer 18

ok, just give me a minute to work it out

Answer 19

\[cotx(\tan ^{4}x+2\tan ^{2}x+1)\] right?

Answer 20

Looks good. Now, distribute \(\cot x\)

Answer 21

should i change the tan and cot to the sin/cos and cos/sin?

Answer 22

Nope, no need for that.

Answer 23

All you need to know is that \(\cot x\) and \(\tan x\) are reciprocals of each other. In other words \((\cot x)(\tan x) = 1\)

Answer 24

But if you need to see that explicitly then

\(\dfrac{\cos x}{\sin x} \dot\ \dfrac{\sin x}{\cos x} = 1\)

Answer 25

so if that =1, would cotx*tan^4 x= tan^3 x?

Answer 26

Correct

Answer 27

THANK YOU!

Answer 28

You're welcome. Good luck with the rest of your work.