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.

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