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cos(tan + cot) = csc only simplify one side to equal csc so far I got this far: [((cos)(sin))/(cos)] + [((cos)(cos))/(sin)] = csc I don't know what to do next

cos(tan + cot) = csc
only simplify one side to equal csc

so far I got this far:
[((cos)(sin))/(cos)] + [((cos)(cos))/(sin)] = csc

I don't know what to do next
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ReportAnswer #1
then on the left..

sin+cos^2/sin=sin+(1-sin^2)sin=sin+csc-sin=csc
answered by: gr8
ReportAnswer #2
You will need a common denominator of sinxcosx in your fraction to get

LS = cosx(sin^2x /(sinxcosx) + cos^2x/(sinxcosx))
= cosx (sin^2x + cos^2x)/(sinxcosx)
= cosx(1)/(sinxcosx)
= 1/sinx
= cscx

BTW you cannot just use the operators, there has to be some "angle" after it

e.g. to say tan = sin/cos is not acceptable and mathematical gibberish.
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cos(tan + cot) = csc only simplify one side to equal csc so far I got this far: [((cos)(sin))/(cos)] + [((cos)(cos))/(sin)] = csc I don't know what to do next
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