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

- anonymous

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

- jamiebookeater

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- anonymous

- SolomonZelman

are you sure it is not tan^2x ?

- anonymous

1 second please let me make sure

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## More answers

- SolomonZelman

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

- anonymous

it is how I initially wrote it

- SolomonZelman

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

- SolomonZelman

\(\large\color{black}{ \displaystyle \cot x\sec^4x=\cot x+2\tan x+\tan^3x }\)
\(\large\color{black}{ \displaystyle \sec^4x=1+2\tan^2x+\tan^4x }\)

- anonymous

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

- SolomonZelman

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

- SolomonZelman

\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\tan x+\tan^3x }\)
\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2x+\tan^4x\right) }\)
would you agree with this?

- SolomonZelman

I factored out of cot(x).

- anonymous

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

- SolomonZelman

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

- SolomonZelman

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

- anonymous

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

- SolomonZelman

well, do you understand why they are logically true?

- dinamix

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

- anonymous

Not really.

- SolomonZelman

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

- SolomonZelman

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

- dinamix

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

- SolomonZelman

turmoilx2, do you agree that:
\(\cot(x) \times \tan(x)=1\)

- anonymous

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

- anonymous

@dinamix . Are you in FLVS?

- SolomonZelman

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

- anonymous

I understand that.

- SolomonZelman

turmolix, how about now?

- anonymous

Yes!

- SolomonZelman

ok, my comp glitched:)

- SolomonZelman

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

- dinamix

@turmoilx2 no mate

- SolomonZelman

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

- anonymous

I understand why the identities are true now

- SolomonZelman

Ok, so now back to our question;)

- SolomonZelman

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

- anonymous

No, please explain.

- SolomonZelman

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?

- SolomonZelman

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

- anonymous

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

- SolomonZelman

(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).

- anonymous

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...

- anonymous

oh wait...

- anonymous

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

- SolomonZelman

yes

- anonymous

I think I understand some more now...

- SolomonZelman

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.

- anonymous

Yes. That would be appreciated, thank you!

- SolomonZelman

\(\large\color{black}{ \displaystyle \cot x \sec^4x=\color{blue}{\cot x}+\color{green}{2\tan x}+\color{red}{\tan^3x} }\)
`-----------------------------------------------`\({\\[0.8em]}\)
\(\large\color{blue}{ \displaystyle \cot x =\cot x }\) (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:
\(\large\color{black}{ \displaystyle \cot x \sec^4x=\color{blue}{\cot x}+\color{green}{2\cot x\tan^2 x}+\color{red}{\cot x\tan^4x} }\)

- SolomonZelman

take your time.

- SolomonZelman

When you are done reading, tell me if you understand how I got the following equation below:
\(\large\color{black}{ \displaystyle \cot x \sec^4x=\color{blue}{\cot x}+\color{green}{2\cot x\tan^2 x}+\color{red}{\cot x\tan^4x} }\)

- anonymous

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???

- anonymous

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

- SolomonZelman

\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\cot x\tan^2 x+\cot x\tan^4x}\)
Rather, take out \(\cot(x)\) from the entire right side.
\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2 x+\tan^4x\right)}\)

- SolomonZelman

did you get everything I have done so far/

- anonymous

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

- SolomonZelman

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

- SolomonZelman

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

- anonymous

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

- SolomonZelman

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

- SolomonZelman

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

- anonymous

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

- SolomonZelman

yes, true

- SolomonZelman

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

- SolomonZelman

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

- anonymous

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

- SolomonZelman

no, that is not true

- SolomonZelman

w²+1 would be w²+1²,
BUT
(w+1)² is different:

- SolomonZelman

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

- SolomonZelman

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

- SolomonZelman

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

- anonymous

I get it now. Mind blown

- anonymous

You are very patient. Thanks

- SolomonZelman

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

- SolomonZelman

By the way, it is not a problem.

- anonymous

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

- SolomonZelman

Yes, that is correct!

- anonymous

THANK YOU SO MUCH!

- SolomonZelman

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

- SolomonZelman

You are welcome:)

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