Question

Verify the trigonometric equation cot(x)sec^4(x)=cot(x)+2tan(x)+tan^3(x)?

Answer 1

@Hero , @ganeshie8 @iambatman help please

Answer 2

are you sure it is not tan^2x ?

Answer 3

1 second please let me make sure

Answer 4

oh, no nvm, it should be like that... but still make sure.

Answer 5

it is how I initially wrote it

Answer 6

I would start from multiplying times tan(x) on both sides.

Answer 7

\(\large\color{black}{ \displaystyle \cot x\sec^4x=\cot x+2\tan x+\tan^3x }\)

\(\large\color{black}{ \displaystyle \sec^4x=1+2\tan^2x+\tan^4x }\)

Answer 8

I have to focus on one side though, I am trying to prove that they are the same, not solve.

Answer 9

oh, okay, it would have been wonderful, but I guess not

Answer 10

\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\tan x+\tan^3x }\)

\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2x+\tan^4x\right) }\)

would you agree with this?

Answer 11

I factored out of cot(x).

Answer 12

I have seen that everywhere I have looked, but I dont understand where the tan^2 or the Tan^4 come from

Answer 13

well, do you agree that:

\(\cot(x) \times \tan(x)=1\)

\(\cot(x) \times \tan^2(x)=\tan(x)\)

\(\cot(x) \times \tan^3(x)=\tan^2(x)\)

\(\cot(x) \times \tan^4(x)=\tan^3(x)\)

and so on...

\(\cot(x) \times \tan^{N}(x)=\tan^{N-1}(x)\)

Answer 14

if you don't understand why these are true, then ask

Answer 15

I didn't know those identities. I knew that sin^2x + cos^2x =1 ....

Answer 16

well, do you understand why they are logically true?

Answer 17

@SolomonZelman i want ask u how did u get sec^4(x) = 1 +2tan^2(x) +tan^4(x)

Answer 18

Not really.

Answer 19

I didn't get that yet, but we will get that.

Answer 20

\(\large\color{black}{ \displaystyle (x+1)^2=x^2+2x+1 }\) right?

Answer 21

we have only sec(x) = 1/cos(x)

Answer 22

turmoilx2, do you agree that:

\(\cot(x) \times \tan(x)=1\)

Answer 23

I suppose... if you tell me that it's true

Answer 24

@dinamix . Are you in FLVS?

Answer 25

\(\cot(x) \times \tan(x)=1\)

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

Answer 26

I understand that.

Answer 27

turmolix, how about now?

Answer 28

Yes!

Answer 29

ok, my comp glitched:)

Answer 30

\(\cot(x) \times \tan(x)=1\)

So, now, we will be multiplying times tan(x) on both sides:

\(\cot(x) \times \tan^2(x)=\tan(x)\)

\(\cot(x) \times \tan^3(x)=\tan^2(x)\)

\(\cot(x) \times \tan^4(x)=\tan^3(x)\)

and so on...

\(\cot(x) \times \tan^{N}(x)=\tan^{N-1}(x)\)
--------------------------------

Is this better now, why the list of these identites is true?

Answer 31

@turmoilx2 no mate

Answer 32

turmoilx, when you read, reply me if you got what I elaborates so far in my last reply.

Answer 33

I understand why the identities are true now

Answer 34

Ok, so now back to our question;)

Answer 35

\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\tan x+\tan^3x }\)

Now, using these identies, I am factoring the left side out of \(\cot(x)\).

\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2x+\tan^4x\right) }\)

Do you get how I am factoring?

Answer 36

No, please explain.

Answer 37

let me do an intermediate step in blue:

\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\tan x+\tan^3x }\)

\(\large\color{blue}{ \displaystyle \cot x \sec^4x=\cot x\cdot 1+2\cot x\tan^2x+\cot x\tan^4x }\)

\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2x+\tan^4x\right) }\)

is it better now?

Answer 38

after the blue, I factor out of \(\cot(x)\).
If you still have questions about how I factored, then please ask.

Answer 39

Im still lost. Sorry if I'm being slow here, online math classes are not for me.

Answer 40

(it is fine if you don't I can attempt to explain a little better than what I did.
I am not a very good tutor after all).

Answer 41

I dont understand how you can just multiply the tan by cot and I dont know how to get the exponents for the tangents still. I understand the identities you mentioned above but I dont know how they apply to this first step
if that makes sense...

Answer 42

oh wait...

Answer 43

Is it allowed because Cotxtanx is the same as multiplying by 1 which wouldnt change the equation?

Answer 44

yes

Answer 45

I think I understand some more now...

Answer 46

but I would still just in case post my explanation on that as throuhg as I can.
I will try not to take too much time.

Answer 47

Yes. That would be appreciated, thank you!

Answer 48

\(\large\color{black}{ \displaystyle \cot x \sec^4x=\color{blue}{\cot x}+\color{green}{2\tan x}+\color{red}{\tan^3x} }\)

`-----------------------------------------------`\({\\[0.8em]}\)
\(\large\color{blue}{ \displaystyle \cot x =\cot x }\) (We know that)\({\\[0.8em]}\)
`-----------------------------------------------`\({\\[0.8em]}\)
\(\large\color{green}{ \displaystyle \tan x =\cot x\tan^2x }\)\({\\[0.8em]}\)
(As we showed before in the list of those identites)\({\\[0.8em]}\)
\(\large\color{green}{ \displaystyle \tan x =\cot x\tan^2x }\)\({\\[0.8em]}\)
And then multiply times 2 on both sides\({\\[0.8em]}\)
\(\large\color{green}{ \displaystyle 2\tan x =2\cot x\tan^2x }\) \({\\[0.8em]}\)
`-----------------------------------------------`\({\\[0.8em]}\)
\(\large\color{red}{ \displaystyle \tan^3x =\cot x\tan^4x }\)\({\\[0.8em]}\)
(Also, as we showed before in the list of those identites)\({\\[0.8em]}\)
`-----------------------------------------------`

So it comes out that:
\(\large\color{black}{ \displaystyle \cot x \sec^4x=\color{blue}{\cot x}+\color{green}{2\cot x\tan^2 x}+\color{red}{\cot x\tan^4x} }\)

Answer 49

take your time.

Answer 50

When you are done reading, tell me if you understand how I got the following equation below:

\(\large\color{black}{ \displaystyle \cot x \sec^4x=\color{blue}{\cot x}+\color{green}{2\cot x\tan^2 x}+\color{red}{\cot x\tan^4x} }\)

Answer 51

Here is what I have so far:
\[CotxSec^4x=Cotx+2Tanx+Tan^3x\]
\[CotxSec^4x=Cotx+2(CotxTan^2x)+(CotxTan^4x)\]
\[CotxSec^4x=Cotx+Cot(2Tan^2x+tan^4x)
Correct???

Answer 52

oops
\[CotxSec^4x=Cotx+Cot(2Tan^2x+tan^4x)\]

Answer 53

\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\cot x\tan^2 x+\cot x\tan^4x}\)

Rather, take out \(\cot(x)\) from the entire right side.

\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2 x+\tan^4x\right)}\)

Answer 54

did you get everything I have done so far/

Answer 55

\[cotxsec4x=cotx((((1))))+2\tan2x+\tan4x)\]
So Im assuming this "1" is from removing the cotagents... but how?

Answer 56

well when I factor cot(x) out of cot(x), then I get 1.

Answer 57

\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\cot x\tan^2 x+\cot x\tan^4x}\)

I am factoring out of \(\cot(x)\). And I factor every term on the left side.

\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2 x+\tan^4x\right)}\)

still questions about that factoring out?

Answer 58

I think what I dont get is why Cotx+2CotxTan^2x doesnt become 1+2Tanx instead of 1+ 2Tan^2x

Answer 59

2tan^2x is what I got, didn't I?

Answer 60

oh, because you have 2cot(x)tan²(x), and thus when you take out the cot(x) part, you remain with 2tan²(x).

Answer 61

I'll just go with it. Cot can be factored from itself to equal 1.

Answer 62

yes, true

Answer 63

\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2 x+\tan^4x\right)}\)

then, you know that: \(\color{red}{({\rm w}+1)^2={\rm w}^2+2{\rm w}+1}\)
So, then look at the parenthesis on the right side
and tell me how THAT can be factors.

Answer 64

\(\large (\)\(\color{red}{\rm w}\) in our case is \(\tan^2x\)\(\large )\)

Answer 65

\[(w+1)^2=w^2+1^2\]
I know that this much is true.

Answer 66

no, that is not true

Answer 67

w²+1 would be w²+1²,
BUT

(w+1)² is different:

Answer 68

\((w+1)^2=(w+1)(w+1)=w(w+1)+1(w+1) \\ =w^2+w+w+1=w^2+2w+1\)

Answer 69

so, therefore we know that:
\((w+1)^2=w^2+2w+1\)

Answer 70

ok, tell me if you get my very last post...

Answer 71

I get it now. Mind blown

Answer 72

You are very patient. Thanks

Answer 73

Ok, so now if you look at, and understand the fact that:
\((w+1)^2=w^2+2w+1\)

Then you should also know that:
\((\tan^2x+1)^2=\left(\tan^2x\right)^2+2\left(\tan^2x\right)+1\)

`-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-` `-`

And then take a look at your problem:
\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2 x+\tan^4x\right)}\)

and it should follow that:
\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+\tan^2 x\right)^2}\)

Answer 74

By the way, it is not a problem.

Answer 75

and then (1+tan^2x)^2 = (sec^2x)^2 which then equals sec^4x

Answer 76

Yes, that is correct!

Answer 77

THANK YOU SO MUCH!

Answer 78

And then, we have verified the original equation to be an identity.

Answer 79

You are welcome:)