Solving for cot2x + sec2x = tan2x + csc2x

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AI Thread Summary
The discussion focuses on solving the equation cot2x + sec2x = tan2x + csc2x. Participants work through the left-hand side (LHS) and right-hand side (RHS) separately, simplifying their expressions. The LHS simplifies to 1 / (sin^2x cos^2x), while the RHS also simplifies to the same expression, confirming both sides are equal. There are clarifications about proper formatting of fractions and the importance of common denominators in the calculations. The conclusion indicates that both sides of the equation are indeed equal, leading to the solution.
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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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