SOLUTION: (cscx+ cotx)(1-cosx)=sinx I don't know how to proove the question. I tried and got a different answer. Can you help me?

Algebra ->  Trigonometry-basics -> SOLUTION: (cscx+ cotx)(1-cosx)=sinx I don't know how to proove the question. I tried and got a different answer. Can you help me?      Log On


   



Question 568765: (cscx+ cotx)(1-cosx)=sinx
I don't know how to proove the question. I tried and got a different answer. Can you help me?

Found 2 solutions by Alan3354, sEahors3:
Answer by Alan3354(69443) About Me  (Show Source):
You can put this solution on YOUR website!
(cscx+ cotx)(1-cosx)=sinx
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Change everything to sines and cosines.

Answer by sEahors3(6) About Me  (Show Source):
You can put this solution on YOUR website!
you change everything to sin and cos .. so what u do is you prove the left side since it will be easier
%281%2Fsinx+%2B+cosx%2Fsinx%29%281-cosx%29+=+sinx
since in the first term we have the same denominator they can be one fraction
so it will be
%28%281%2Bcosx%29%2Fsinx%29%281-cosx%29+=+sinx
now we just multiply the two terms
%28%281%2Bcosx%29%281-cosx%29%29%2Fsinx+=+sinx
now we know when we have two terms being multiplied and the sign is different we will have something like this
%28x-y%29%28x%2By%29+=+x%5E2-y%5E2
same thing applies for the numerator in this equation
%281%2Bcosx%29%281-cosx%29%2Fsinx+=+sinx
%28%281%5E2-cosx%5E2%29%2Fsinx%29+=+sinx
and we also know the trigonometric identity that says
sinx%5E2+=1-cosx%5E2
so we substitute that in for the numerator also and we get
sinx%5E2%2Fsinx=sinx
dividing two same number with different exponents means subtracting the exponent so we will end up with
sinx+=+sinx

note : it is supposed to be the sin that is squared .. not the X .. but the website didnt allow me to do that so =/ ... but i hope this helps =)