# Help with Trigonometric Identities

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• February 15th 2011, 08:43 PM
toeknee
Help with Trigonometric Identities
$sec^2{x}+tan^2{x}sec^2{x}=sec^4{x}$

So far I have changed everything to fractions using the quotient properties, but I have no idea where to go next.
• February 15th 2011, 08:46 PM
Prove It
Use the identity $\displaystyle \tan^2{x} + 1 = \sec^2{x} \implies \tan^2{x} = \sec^2{x} - 1$.

Substitute this into the LHS and simplify.
• February 15th 2011, 08:57 PM
toeknee
Sorry, but what does LHS stand for?
• February 15th 2011, 09:02 PM
topsquark
Quote:

Originally Posted by toeknee
Sorry, but what does LHS stand for?

Left Hand Side. Meaing if we have the expression a + 1 = b then the LHS is a + 1.

-Dan
• February 15th 2011, 09:02 PM
Prove It
Left Hand Side
• February 15th 2011, 09:10 PM
toeknee
You can still use the Pythagorean property for $tan^2{x}sec^2{x}$?
• February 15th 2011, 09:14 PM
Prove It
Surely $\displaystyle \tan^2{x}\sec^2{x} = (\sec^2{x} - 1)\sec^2{x}$...
• February 15th 2011, 09:18 PM
toeknee
I thought because it was one term, you couldn't, but I guess not. Thanks a lot for your help. Kind of fitting that your name is Prove It.