# Verify Trig expression

• December 5th 2010, 01:41 PM
IDontunderstand
Verify Trig expression
I wish I knew how to use the Latex:

Sec^2(x)Csc^2(x)=Sec^2(x)+Cot^2(x)

I have tried to work with both sides of the expression at different times and looked at the work on each side to determine if I could get the solution,

I have left the left side alone and changed the right side to be

= Sec^2(x) + (csc^2x-1) But I get stuck.
Any help with the side and which I should change would be great.

Thanks
• December 5th 2010, 01:46 PM
DrSteve
I would suggest trying the following:

Rewrite the right-hand side in terms of sine and cosine.

Get a common denominator.

Use a pythagorean identity to write the numerator in terms of just sine OR just cosine.

Factor the numerator, and do some cancellation.

Rewrite in terms of sec and csc

(I haven't actually written out the details but this procedure should work).
• December 5th 2010, 01:53 PM
DrSteve
I think you may have a typo in your question. Are you sure that the Cot on the right shouldn't be Csc?
• December 5th 2010, 02:02 PM
Quote:

Originally Posted by IDontunderstand
I wish I knew how to use the Latex:

Hit the TEX and write the math between the "MATH" brackets..

$Sec^2(x)Csc^2(x)=Sec^2(x)+Cot^2(x)$

I have tried to work with both sides of the expression at different times and looked at the work on each side to determine if I could get the solution,

I have left the left side alone and changed the right side to be

$= Sec^2(x) + \left(csc^2x-1\right)$

But I get stuck.
Any help with the side and which I should change would be great.

Thanks

$sec^2x\;csc^2x=sec^2x+csc^2x-1$

Bring the $sec^2x$ terms together...

$sec^2x\;csc^2x-sec^2x=csc^2x-1$

Factor...

$sec^2x\left(csc^2x-1\right)=csc^2x-1$

which is solvable if you started with an equation and are finding x,
but not if you want both sides equal irrespective of x.
• December 5th 2010, 02:05 PM
IDontunderstand
Thank you very much. The Cot needs to be a CSC. That could be why I cannot get it to work out. I looked at the problem a lot of times I guess I saw what I wanted. Wow imagine it works when the problem is correct. Its funny how that happens.