Question

I need some help solving this identity:

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

Answer 1

what do you mean ?

Answer 2

"Verify each trigonometric equation by substituting identities to match the right hand side of the equation to the left hand side of the equation."

These are the exact directions. I guess I have to make the right look the like the left.

Answer 3

first convert every term to sines and cosines

\[\tan\theta=\tfrac{\sin x}{\cos x}\]\[\csc\theta=\tfrac1{\sin\theta}\]\[\sec\theta=\tfrac1{\cos\theta}\]\[\cot\theta=\tfrac1{\tan\theta}=\tfrac{\cos\theta}{\sin\theta }\]

Answer 4

I did that and that's where I got lost and/or made a mistake.

Answer 5

what did you get for the left hand side?

Answer 6

(1/tanx) + (1+tan2x)(sec2x)

Answer 7

huh ?

Answer 8

that's what I got, so now you see my problem.

Answer 9

the left hand side
cot x * sec^4x

becomes

cos x /sin x * 1/ cos^4 x

simplify this further

Answer 10

wouldn't it then become:

cosx/sinx * 1/(cos^2x)^2

then that would become

cosx/sinx * 1/(1-sin^2)^2

??

Answer 11

why would you do that ?

Answer 12

I have no idea what I'm doing.

Answer 13

Can you start from the beginning and show me step by step?

Answer 14

cot x sec^4x

[cot = 1/ tan = cos / sin] [sec = 1/ cos]

= cos x /sin x * 1/ cos^4 x

Answer 15

okay now what about the other side? How do I make them match?

Answer 16

first simplify

cos x /sin x * 1/ cos^4 x

a little

Answer 17

[one of the cosines cancel]

Answer 18

would it be sinx * 1/cos^3x

Answer 19

1/ sin x * cos^3 x

Answer 20

Now look at the right hand side of the equation


cot x + 2 tan x + tan^3x

what is this in terms of sines and cosines ?

Answer 21

Okay...

cosx/sinx + 2(sinx/cosx) and I don't know how to simplify tan^3x

Answer 22

tan^3 x = sin^3 x / cos^3 x

Answer 23

so the right hand side is

cosx/sinx + 2(sinx/cosx) + sin^3 x / cos^3 x

Answer 24

now compare the left with the right


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

Answer 25

multiply both sides of the equation by sin x

Answer 26

is it:
sin^2x * sinx/cos^3x = sinxcosx/sinx + 2(sin^2x/cosx) + sin^4x/cos^3x
??

Answer 27

almost, the sine cancels on the left to get

1/cos^3 x = sin x * cos x/sin x + 2(sin^2 x/cos x) + sin^4 x / cos^3 x

the sine also cancels on the first term on the right hand side

Answer 28

how does the sin^2x and sinx on the right cancel? Wouldn't I still have a sinx left?

Answer 29

i dont know where you got the sin^2x from

Answer 30

I meant on the left side?

Answer 31

Okay so I now have

cos^3x = cosx + 2(sin^2 x/cos x) + sin^4 x / cos^3 x

?

Answer 32

the term on the left should be 1/cos^3x

Answer 33

so to make the left hand side simplify to 1 , multiply both sides of the equation by cos^3 x

Answer 34

1= cos^4x + 2(cos^3xsin^2x / cosx) + cos^3sin^4x/cos^3)


then 1= cos^4x + 2 (cos^2xsin^2x) + sin^4x

Answer 35

is that right?

Answer 36

yes that's right good work,

so now look at this left hand side

cos^4x + 2 (cos^2xsin^2x) + sin^4x

this can be simplified by factorising

to make this easier to see you could
temporality sub A = cos^2 x, and B = sin^2 x

which gives
A^2 + 2AB + B^2

can you simplify this?