
are we solving ?
cotxcot2xcot2xcot3xcot3xcotx=1
(cosx/sinx)(cos2x/sin2x)  (cos2x/sin2x)(cos3x/sin3x)  (cos3x/sin3x)(cosx/sinx) = 1
common denominator is sinx(sin2x)(sin3x), so
[ cosxcos2xsin3x  sinxcos2xcos3x  sin2xcos3xcosx ] / (sinxsin2xsin3x) = 1
crossmultiply, then bring sinxsin2xsin3x back to left side
cosxcos2xsin3x  sinxcos2xcos3x  sin2xcos3xcosx  sinxsin2xsin3x = 0
factor sin3x from 1st and 4th, cos3 from 2nd and 3rd
sin3x(cosxcos2x  sinxsin2x)  cos3x(sinxcos2x + cosxsin2x) = 0
sin3x(cos(x+2x))  cos3x(sin(x+2x)) = 0
sin3xcos3x  cos3xsin3x = 0
0 = 0
Since this is a true statement, the original equation must have been an identity.
posted by Reiny
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