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Trigonometry 6 Online
OpenStudy (anonymous):

Simplify sine squared theta divided by the quantity 1 minus sine squared theta.

OpenStudy (anonymous):

@iPwnBunnies

OpenStudy (anonymous):

= sin^2 theta/cos^2 theta=tan^2 theta

OpenStudy (anonymous):

\[\frac{ Sin^2(\theta) }{ 1-\ Sin^2(\theta) }\]

OpenStudy (anonymous):

1 tan2 È sin2 È cos2 È those are the answers

OpenStudy (anonymous):

You can use the identity:\[Cos^2(\theta) +Sin^2(\theta) = 1\]

OpenStudy (anonymous):

so 1?

OpenStudy (anonymous):

yes...sure...its 1

OpenStudy (anonymous):

is it 1 bc sin theta cancels out

OpenStudy (anonymous):

Solving for Cos^2(theta) you see:\[\cos^2(\theta) = 1-\sin^2(\theta)\]

OpenStudy (anonymous):

No it is not 1.

OpenStudy (anonymous):

Replace your new identity and you will find your solution.

OpenStudy (anonymous):

its tan^2(e) right?

OpenStudy (anonymous):

You have:\[\frac{ Sin^2(\theta) }{ 1-Sin^2(\theta) }\] And know:\[Cos^2(\theta) =1 - Sin^2(\theta)\] So replacing we now have: \[\frac{ Cos^2(\theta) }{ Sin^2(\theta) }\]

OpenStudy (anonymous):

Another identity can now be used since we have similar exponents: \[\frac{ Cos^n(\theta) }{ Sin^n(\theta) }=Tan^n(\theta)\] Where n = exponent value

OpenStudy (anonymous):

Correct

OpenStudy (anonymous):

so i was right? tan^2 E

OpenStudy (anonymous):

yay thanks! can you help with some more?

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