Solving for cot2x + sec2x = tan2x + csc2x

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Homework Help Overview

The problem involves the equation cot2x + sec2x = tan2x + csc2x, which is situated within the context of trigonometric identities and simplifications.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants are attempting to simplify both sides of the equation, with some focusing on the left-hand side (LHS) and others on the right-hand side (RHS). There are questions regarding the correctness of the transformations and simplifications made during the process.

Discussion Status

Some participants have provided feedback on the attempts made, suggesting that comparisons between the LHS and RHS should be drawn. There is an ongoing exploration of the expressions derived from both sides, with indications of confusion regarding specific steps in the simplification process.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the extent of guidance provided. There are indications of uncertainty regarding the legality of certain algebraic manipulations and the need for clarity in notation.

Veronica_Oles
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Homework Statement



cot2x + sec2x = tan2x + csc2x

Homework Equations

The Attempt at a Solution


I began by working on my left side.

I got up until
=cos^4x + sin^2x / sin^2x cos^2x

And unsure of where to go next.
 
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Veronica_Oles said:

Homework Statement



cot2x + sec2x = tan2x + csc2x

Homework Equations

The Attempt at a Solution


I began by working on my left side.

I got up until
=cos^4x + sin^2x / sin^2x cos^2x

And unsure of where to go next.
You made a good start working on the LHS. Do the same to the RHS and see what you get.
 
SteamKing said:
You made a good start working on the LHS. Do the same to the RHS and see what you get.
On right side I got sin^4x + cos^2x / cos^2x sin^2x , I can't seem to get it
 
Veronica_Oles said:
On right side I got sin^4x + cos^2x / cos^2x sin^2x , I can't seem to get it
Compare the LHS to the RHS now. Do you notice anything in common?
 
SteamKing said:
Compare the LHS to the RHS now. Do you notice anything in common?
L.S = cos^4x / sin^2x cos^2x + sin^2x / sin^2x cos^2x
= cos^2x / sin^2x + 1 / cos^2x
= cos^2x + sin^2x / sin^2x cos^2x
= 1 / sin^2x cos^2x

R.S = sin^4x / sin^2x cos^2x + cos^2x / sin^2x cos^2x
= sin^2x / cos^2x + 1 / sin^2x
= sin^2x + cos^2x / sin^2x cos^2x
= 1 / sin^2x cos^2x
Would this be the answer?
 
Veronica_Oles said:
L.S = cos^4x / sin^2x cos^2x + sin^2x / sin^2x cos^2x
= cos^2x / sin^2x + 1 / cos^2x
I'm having a problem seeing how you go from the expression above to the one below:
= cos^2x + sin^2x / sin^2x cos^2x
I see that you found the common denominator, but your numerator is incorrect.
= 1 / sin^2x cos^2x

R.S = sin^4x / sin^2x cos^2x + cos^2x / sin^2x cos^2x
= sin^2x / cos^2x + 1 / sin^2x
= sin^2x + cos^2x / sin^2x cos^2x
= 1 / sin^2x cos^2x
Would this be the answer?
Same comments from the LHS calculations apply above.

Compare the expression you found on the RHS here with the one you obtained in Post #3:

Veronica_Oles said:
On right side I got sin^4x + cos^2x / cos^2x sin^2x
 
Veronica_Oles said:
L.S = cos^4x / (sin^2x cos^2x) + sin^2x / (sin^2x cos^2x)
= cos^2x / sin^2x + 1 / cos^2x
= ( ? × cos^2x + sin^2x) / (sin^2x cos^2x)
= 1 / (sin^2x cos^2x)

R.S = sin^4x / (sin^2x cos^2x) + cos^2x / (sin^2x cos^2x)
= sin^2x / cos^2x + 1 / sin^2x ( No idea what you did here. Whatever, it's not legal.)
= (sin^2x + cos^2x) / (sin^2x cos^2x)
= 1 / (sin^2x cos^2x)
Would this be the answer?
When writing "fractions" with an "in-line" format, one using the " / " character, you need to use parentheses to include (the entire numerator) / (the entire denominator).

I have added parentheses to what I assume are the proper locations in the above included quote. (Also added a few other items.)
 

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