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Mathematics 23 Online
OpenStudy (anonymous):

simplify : sin^2x/sec^2x-1

OpenStudy (joannablackwelder):

sec^2x-1=tan^2x

OpenStudy (anonymous):

ik secis 1/cos so I can flip with squareing but idk if im right I think the answer is sin^2xtan^2x I guess im wrong lol

OpenStudy (joannablackwelder):

So I get sin^2x/tan^2x. Since tanx = sinx/cosx, tan^2x = sin^2x/cos^2x

OpenStudy (joannablackwelder):

sin^2x/(sin^2x/cos^2x)= cos^2x

OpenStudy (alekos):

is this sin^2x/(sec^2-1)?

OpenStudy (alekos):

sorry i meant is sec^2x - 1 on the bottom line?

OpenStudy (joannablackwelder):

That is what I am assuming, but I could be wrong. @cohesivesandman ?

OpenStudy (anonymous):

sec^2 x -1

OpenStudy (joannablackwelder):

All of that is in the denominator?

OpenStudy (alekos):

ok, then joanna is correct

OpenStudy (anonymous):

yes

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