Solving for cot2x + sec2x = tan2x + csc2x

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SUMMARY

The equation cot2x + sec2x = tan2x + csc2x can be simplified by manipulating both sides. The left-hand side (LHS) simplifies to 1 / (sin^2x cos^2x) after combining terms, while the right-hand side (RHS) also simplifies to 1 / (sin^2x cos^2x). Both sides are equal, confirming the identity. The key steps involve finding a common denominator and correctly simplifying the fractions.

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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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