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How to verify the trig identity sec^(2)x/secx-1=[csc^(2)x](secx-1)?I've tried...

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sputnik77 | Student, Grade 10 | eNotes Newbie

Posted May 14, 2012 at 10:11 PM via web

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How to verify the trig identity sec^(2)x/secx-1=[csc^(2)x](secx-1)?

I've tried converting the secants to 1/cosines but that doesn't seem to be getting me anywhere.  Thanks for your help!

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rcmath | High School Teacher | (Level 1) Associate Educator

Posted May 14, 2012 at 10:49 PM (Answer #1)

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I think there is an error in your problem the LHS should be +1 not -1

***RHS=`(csc^2x)(secx-1)=1/(sin^2x)(1/(cosx)-1)=`

`1/(sin^2x)((1-cosx)/cosx)=1/(1-cos^2x)((1-cosx)/cosx)=`

`(1-cosx)/((1-cos^2x)*cosx)=1/((1+cosx)*cosx)` 


***LHS `(sec^2x)/(secx+1)=(1/(cos^2x))/(1/cosx+1)=`

`(1/(cos^2x))/((1+cosx)/cosx)=1/(cos^2x)*cosx/(1+cosx)=`

`1/(cosx*(1+cosx))`

Hence LHS=RHS

 

 

 

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sputnik77 | Student, Grade 10 | eNotes Newbie

Posted May 14, 2012 at 11:25 PM (Answer #2)

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I'm pretty sure that the equation should look like this:

 

sec²x/secx-1 = csc²x(secx+1)

I think you switched the signs on each side of the equation, but I might not have written the equation down correctly.  Thanks for your help!

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sputnik77 | Student, Grade 10 | eNotes Newbie

Posted May 14, 2012 at 11:29 PM (Answer #3)

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I just realized that I originally posted the original equation wrong!  The right one is the one that I just posted.  Oops!

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rcmath | High School Teacher | (Level 1) Associate Educator

Posted May 15, 2012 at 2:33 AM (Answer #4)

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If you repeat the above mentioned steps, change the -1 into +1 in the RHS and +1 to -1 in the LHS, you should get the solution to yor problem.

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