Help with Trigonometric Identities

**toeknee** · February 15th 2011, 08:43 PM

$sec^2{x}+tan^2{x}sec^2{x}=sec^4{x}$

So far I have changed everything to fractions using the quotient properties, but I have no idea where to go next.

**Prove It** · February 15th 2011, 08:46 PM

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.

**toeknee** · February 15th 2011, 08:57 PM

Sorry, but what does LHS stand for?

**topsquark** · February 15th 2011, 09:02 PM

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

**Prove It** · February 15th 2011, 09:02 PM

Left Hand Side

**toeknee** · February 15th 2011, 09:10 PM

You can still use the Pythagorean property for $tan^2{x}sec^2{x}$ ?

**Prove It** · February 15th 2011, 09:14 PM

Surely $\displaystyle \tan^2{x}\sec^2{x} = (\sec^2{x} - 1)\sec^2{x}$ ...

**toeknee** · February 15th 2011, 09:18 PM

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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