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Mathematics 18 Online
OpenStudy (anonymous):

How do I prove the identity: 1/sinx - sinx = cosx/tanx?

OpenStudy (saifoo.khan):

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

OpenStudy (anonymous):

= 1 - sin^2 x ----------- yea saifoo got it sin x

OpenStudy (saifoo.khan):

\[\frac{\cos x * \cos x}{\sin x} \to \cos x* \frac{\cos x}{\sin x} \to \cos x (\frac{1}{\tan}) \]

OpenStudy (anonymous):

wait why is did you multiply sinx again? ahh D:

OpenStudy (saifoo.khan):

Where?

OpenStudy (anonymous):

1- sinx*sinx

OpenStudy (saifoo.khan):

I took LCM as sinx

OpenStudy (anonymous):

the lowest common multiple?

OpenStudy (saifoo.khan):

Yes sir.

OpenStudy (anonymous):

okay thanks!

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