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OpenStudy (anonymous):

(sin2x)/(1+cos2x)+(1+cos2x)/(sin2x)=2/(sin2x) simplify..

OpenStudy (anonymous):

have any idea?:)

OpenStudy (anonymous):

maybe start with \[1+\cos(2x)=2\cos^2(x)\]

OpenStudy (anonymous):

why 1+cos(2x) = 2cos^2(x)??

OpenStudy (anonymous):

why is it true or why use it?

OpenStudy (anonymous):

why it is true..?

OpenStudy (anonymous):

because \[\cos(2x)=2\cos^2(x)-1\]

OpenStudy (anonymous):

also \[\sin(2x)=2\cos(x)\sin(x)\] use these and you will probably get the answer quickly

OpenStudy (anonymous):

or maybe just add up and see what we get

OpenStudy (anonymous):

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

OpenStudy (anonymous):

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

OpenStudy (anonymous):

thanks :))

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