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Mathematics 19 Online
OpenStudy (anonymous):

i cant verify 1+cos2x / sin2x = cotx....HELP

OpenStudy (anonymous):

it should be (1+cos2x)/sin2x

OpenStudy (anonymous):

\[1+\cos2x=2 \cos ^{2}x\] and sin2x =2 sinx cosx

OpenStudy (anonymous):

\[\frac{ 1+\cos2x }{ \sin2x } = cotx\]

OpenStudy (anonymous):

\[1+\cos2x=2\cos ^{2}x\] and sin2x=2sinxcosx substitute these two to get the answer

OpenStudy (anonymous):

cosx/sinx=cotx

OpenStudy (anonymous):

so it becomes like \[\frac{ 2\cos^2x }{ 2sinxcosx }\]

OpenStudy (anonymous):

???

OpenStudy (anonymous):

but what do i do after that?

OpenStudy (anonymous):

cancel the common terms. then cosx/sinx=cotx

OpenStudy (anonymous):

so it then becomes 2cosx / 2sinx

OpenStudy (anonymous):

and the twos cancels out...so it becomes cosx / sinx!! thank you!

OpenStudy (anonymous):

ok! remember to close the question guys and if appropriate, to award medals:)

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