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ShadowysBest ResponseYou've already chosen the best response.0
I believe you have attempted to do it. Where were you stuck? Though the answer isn't that pretty..
 one year ago

SugarRainbowBest ResponseYou've already chosen the best response.0
okay so i was thinking of multiplying cosx to the numerator and denominator of the (sinx/1cosx) so then the denominator will be sin^2x and then multiply sinx to (1cosx/sinx) so that both denominators will bbe sin^2x is that even a good first step to take?
 one year ago

seitysBest ResponseYou've already chosen the best response.1
sinx/(1cosx)*(1+cosx)/(1+cosx) + (1cosx/sinx) =sinx(1+cosx)/sin^2 x + (1cosx/sinx) =(1+cosx/sinx)+(1cosx/sinx) =(1+cosx+1cosx)/sinx =2/sinx
 one year ago

SugarRainbowBest ResponseYou've already chosen the best response.0
why did you do this sinx(1+cosx)/sin^2 x or how'd that happen
 one year ago

SugarRainbowBest ResponseYou've already chosen the best response.0
oh wait nevermind i see what you did there
 one year ago

seitysBest ResponseYou've already chosen the best response.1
when you multiply (1cosx) by its conjugate (1+cosx), you get (1cos^2 x) which is just sin^2 x > sin^2 x + cos^2 x = 1 so 1cos^2 x=sin^2 x
 one year ago

SugarRainbowBest ResponseYou've already chosen the best response.0
okay well how did that turn into (1+cosx/sinx)?
 one year ago

seitysBest ResponseYou've already chosen the best response.1
there is a sinx in the numerator and a sin^2 x in the denominator. so 1 of them cancels out.
 one year ago

SugarRainbowBest ResponseYou've already chosen the best response.0
and you got 2/sinx because the +cosx and the cosx cancel out right?
 one year ago

seitysBest ResponseYou've already chosen the best response.1
because that is the remaining denominator for both terms.
 one year ago

seitysBest ResponseYou've already chosen the best response.1
you can simplify it even further because 1/sinx = cscx so it can also be written as 2cscx
 one year ago

SugarRainbowBest ResponseYou've already chosen the best response.0
oh okay thank you this was really helpful
 one year ago
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