then on the left..
sin+cos^2/sin=sin+(1-sin^2)sin=sin+csc-sin=csc
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
2 answers
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.
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.