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Precalculus 8 Online
OpenStudy (anonymous):

Graph the expression sec^ 2 x - sec^ 2 x sin^ 2 x on your calculator. Determine what constant or single function is equivalent to the given expression

OpenStudy (anonymous):

Factor out the \(\sec^2x\).

OpenStudy (anonymous):

I did I got sin ^2x as whats left

OpenStudy (psymon):

Not quite :P

OpenStudy (anonymous):

\[\sec^2x-\sec^2x\sin^2x=\sec^2x\left(1-\sin^2x \right)\]

OpenStudy (anonymous):

Ok so it would be\[\sec ^2x - \sin^2x\]

OpenStudy (psymon):

Siths just wrote the answer you should have :P Might as well copy it, it is correct.

OpenStudy (psymon):

\[\sec ^{2}x(1-\sin ^{2}x)\] If you multiplied it back in you would get the same answer you started with :3

OpenStudy (anonymous):

Ohhhhhhhh ok thanks guys can you help me on the ones I posted above this one I swear it will be the last one

OpenStudy (psymon):

Whoa, whoa, we're not done.

OpenStudy (anonymous):

Oh what

OpenStudy (psymon):

You can simplify what ya got there more. Use the fact that: \[\sin ^{2}x+\cos ^{2}x=1\] to rewrite that 1-sin^2(x) part.

OpenStudy (anonymous):

So would this be left \[\sec ^2x \times \cos^2x\]

OpenStudy (psymon):

Correct. Now what's the one last thing you can do to simplify that? :P

OpenStudy (anonymous):

\[\frac{ 1 }{ \cos^2x } \times \cos^2x \]

OpenStudy (psymon):

There ya go. Therefore: \[f(x)=1\]

OpenStudy (psymon):

So all that mumbo jumbo was just a funky way of writing the line y = 1, lol.

OpenStudy (anonymous):

Yeah finally can you help me with my last problem with explanations

OpenStudy (psymon):

What is it?

OpenStudy (anonymous):

I will post it as a new question hold on

OpenStudy (psymon):

kk

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