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Mathematics 17 Online
OpenStudy (anonymous):

PROV THE IDENTITY TAN^2Θ= ((1-cos(2Θ))/((1+cos(2Θ))

OpenStudy (anonymous):

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OpenStudy (anonymous):

@Mertsj can you tell me how to solve this

OpenStudy (anonymous):

anyone?

OpenStudy (helder_edwin):

try using \[\large cos(2x)=\cos^2x-\sin^2x \] it will follow easyly from this

OpenStudy (mertsj):

Going to use x instead of theta

OpenStudy (anonymous):

ok

OpenStudy (mertsj):

\[\frac{1-\cos 2x}{1+\cos 2x}=\frac{1-(\cos ^2x-\sin ^2x)}{1+(\cos^2-\sin ^2x)}=\frac{1-\cos ^2x+\sin ^2x}{1-\sin ^2x+\cos ^2x}\]

OpenStudy (anonymous):

1st of all i used the wrong identity.. ok i get that part now though

OpenStudy (mertsj):

\[\frac{\sin ^2x+\sin ^2x}{\cos ^2x+\cos ^2x}=\frac{2\sin ^2x}{2\cos ^2x}\]

OpenStudy (anonymous):

okay. i understand that too

OpenStudy (anonymous):

sin^2 is 1-cos^2 right? so would you do that next?

OpenStudy (mertsj):

What is 2/2 ?

OpenStudy (anonymous):

1. ok so it cancels

OpenStudy (mertsj):

What is: \[\frac{\sin x}{\cos x}\]

OpenStudy (anonymous):

ohhh i understand. so youre left w/ sin^2x/ cos^2x which is tan^2x

OpenStudy (mertsj):

yes

OpenStudy (anonymous):

thank you

OpenStudy (anonymous):

can you help me w/ one more

OpenStudy (anonymous):

i can post it as new question

OpenStudy (mertsj):

possibly

OpenStudy (mertsj):

What is it?

OpenStudy (anonymous):

its another proving trig identity

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