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

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@Hero , @ganeshie8 @iambatman help please

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5are you sure it is not tan^2x ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01 second please let me make sure

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5oh, no nvm, it should be like that... but still make sure.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0it is how I initially wrote it

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5I would start from multiplying times tan(x) on both sides.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5\(\large\color{black}{ \displaystyle \cot x\sec^4x=\cot x+2\tan x+\tan^3x }\) \(\large\color{black}{ \displaystyle \sec^4x=1+2\tan^2x+\tan^4x }\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have to focus on one side though, I am trying to prove that they are the same, not solve.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5oh, okay, it would have been wonderful, but I guess not

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\tan x+\tan^3x }\) \(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2x+\tan^4x\right) }\) would you agree with this?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5I factored out of cot(x).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I have seen that everywhere I have looked, but I dont understand where the tan^2 or the Tan^4 come from

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5well, 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... \(\cot(x) \times \tan^{N}(x)=\tan^{N1}(x)\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5if you don't understand why these are true, then ask

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I didn't know those identities. I knew that sin^2x + cos^2x =1 ....

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5well, do you understand why they are logically true?

dinamix
 one year ago
Best ResponseYou've already chosen the best response.0@SolomonZelman i want ask u how did u get sec^4(x) = 1 +2tan^2(x) +tan^4(x)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5I didn't get that yet, but we will get that.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5\(\large\color{black}{ \displaystyle (x+1)^2=x^2+2x+1 }\) right?

dinamix
 one year ago
Best ResponseYou've already chosen the best response.0we have only sec(x) = 1/cos(x)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5turmoilx2, do you agree that: \(\cot(x) \times \tan(x)=1\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I suppose... if you tell me that it's true

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@dinamix . Are you in FLVS?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5\(\cot(x) \times \tan(x)=1\) \(\dfrac{\cos(x)}{\sin(x)}\times \dfrac{\sin(x)}{\cos(x)}=1\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5turmolix, how about now?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5ok, my comp glitched:)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5\(\cot(x) \times \tan(x)=1\) 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... \(\cot(x) \times \tan^{N}(x)=\tan^{N1}(x)\)  Is this better now, why the list of these identites is true?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5turmoilx, when you read, reply me if you got what I elaborates so far in my last reply.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I understand why the identities are true now

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5Ok, so now back to our question;)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\tan x+\tan^3x }\) 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?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5let 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
 one year ago
Best ResponseYou've already chosen the best response.5after the blue, I factor out of \(\cot(x)\). If you still have questions about how I factored, then please ask.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Im still lost. Sorry if I'm being slow here, online math classes are not for me.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5(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
 one year ago
Best ResponseYou've already chosen the best response.0I 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
 one year ago
Best ResponseYou've already chosen the best response.0Is it allowed because Cotxtanx is the same as multiplying by 1 which wouldnt change the equation?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I think I understand some more now...

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5but 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
 one year ago
Best ResponseYou've already chosen the best response.0Yes. That would be appreciated, thank you!

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5\(\large\color{black}{ \displaystyle \cot x \sec^4x=\color{blue}{\cot x}+\color{green}{2\tan x}+\color{red}{\tan^3x} }\) ``\({\\[0.8em]}\) \(\large\color{blue}{ \displaystyle \cot x =\cot x }\) (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
 one year ago
Best ResponseYou've already chosen the best response.5When 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
 one year ago
Best ResponseYou've already chosen the best response.0Here 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
 one year ago
Best ResponseYou've already chosen the best response.0oops \[CotxSec^4x=Cotx+Cot(2Tan^2x+tan^4x)\]

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\cot x\tan^2 x+\cot x\tan^4x}\) 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
 one year ago
Best ResponseYou've already chosen the best response.5did you get everything I have done so far/

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[cotxsec4x=cotx((((1))))+2\tan2x+\tan4x)\] So Im assuming this "1" is from removing the cotagents... but how?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5well when I factor cot(x) out of cot(x), then I get 1.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x+2\cot x\tan^2 x+\cot x\tan^4x}\) 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
 one year ago
Best ResponseYou've already chosen the best response.0I think what I dont get is why Cotx+2CotxTan^2x doesnt become 1+2Tanx instead of 1+ 2Tan^2x

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.52tan^2x is what I got, didn't I?

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5oh, because you have 2cot(x)tan²(x), and thus when you take out the cot(x) part, you remain with 2tan²(x).

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I'll just go with it. Cot can be factored from itself to equal 1.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5\(\large\color{black}{ \displaystyle \cot x \sec^4x=\cot x\left(1+2\tan^2 x+\tan^4x\right)}\) 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
 one year ago
Best ResponseYou've already chosen the best response.5\(\large (\)\(\color{red}{\rm w}\) in our case is \(\tan^2x\)\(\large )\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[(w+1)^2=w^2+1^2\] I know that this much is true.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5no, that is not true

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5w²+1 would be w²+1², BUT (w+1)² is different:

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5\((w+1)^2=(w+1)(w+1)=w(w+1)+1(w+1) \\ =w^2+w+w+1=w^2+2w+1\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5so, therefore we know that: \((w+1)^2=w^2+2w+1\)

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5ok, tell me if you get my very last post...

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I get it now. Mind blown

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You are very patient. Thanks

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5Ok, 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
 one year ago
Best ResponseYou've already chosen the best response.5By the way, it is not a problem.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and then (1+tan^2x)^2 = (sec^2x)^2 which then equals sec^4x

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5Yes, that is correct!

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5And then, we have verified the original equation to be an identity.

SolomonZelman
 one year ago
Best ResponseYou've already chosen the best response.5You are welcome:)
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