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1-sec8x/1-sec4x=tan8x/tan4x

Guest Mar 5, 2017
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3+0 Answers

 #1
avatar+91259 
0

There is no brackets so I am not really sure what you want.

 

1-sec8x/1-sec4x=tan8x/tan4x

 

Are you asking for all real values of x given that:   ??

I suppose that is what you want.

 

\(\frac{1-sec(8x)}{1-sec(4x)}=\frac{tan(8x)}{tan(4x)}\\ \)

 

 

https://www.wolframalpha.com/input/?i=((1-sec(8x))%2F(1-(sec(4x))))%3Dtan(8x)%2Ftan(4x)

 

I keep getting a different answer but I'd trust wolfram|alpha over me.

Melody  Mar 5, 2017
 #2
avatar+79894 
0

This is not an identity for  either

 

[1-sec 8x] /[1- sec4x]  = tan8x / tan4x      nor

 

(1 - sec^8(x)) / (1 - sec^4(x))  = tan^8(x) / tan^4(x)

 

See here : https://www.desmos.com/calculator/figbjj45d8 

 

And  here : https://www.desmos.com/calculator/mbarrthbsp

 

 

cool cool cool

CPhill  Mar 5, 2017
 #3
avatar+91259 
0

Hi Chris,

No it is definitely not an identity.  The question does not say 'prove'

I just took it as a solve for x question.  :)

Melody  Mar 6, 2017

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