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Question Number 105916 by bobhans last updated on 02/Aug/20
prove that cot x. cot 2x +cot 2x. cot 3x+2 = cot x.(cot x−cot 3x)
provethatcotx.cot2x+cot2x.cot3x+2=cotx.(cotx−cot3x)
Answered by bemath last updated on 01/Aug/20
cot x cot 2x + cot 2x cot 3x = cot 2x (cot x+cot 3x)= (((cot^2 x−1)/(2cot x))) (cot x+((cot^3 x−3cot x)/(3cot^2 x−1)))= ((cot^2 x−1)/(2cot x)) .((3cot^3 x−cot x+cot^3 x−3cot x)/(3cot^2 x−1)) ((cot^2 x−1)/(2cot x)) .((4cot^3 x−4cot x)/(3cot^2 x−1))= (cot^2 x−1).((2cot^2 x−2)/(3cot^2 x−1)) = ((2cot^3 x−2cot^2 x−2cot^2 x+2)/(3cot^2 x−1)) = ((2cot^3 x−4cot^2 x+2)/(3cot^2 x−1)).
cotxcot2x+cot2xcot3x=cot2x(cotx+cot3x)=(cot2x−12cotx)(cotx+cot3x−3cotx3cot2x−1)=cot2x−12cotx.3cot3x−cotx+cot3x−3cotx3cot2x−1cot2x−12cotx.4cot3x−4cotx3cot2x−1=(cot2x−1).2cot2x−23cot2x−1=2cot3x−2cot2x−2cot2x+23cot2x−1=2cot3x−4cot2x+23cot2x−1.
Answered by bobhans last updated on 02/Aug/20
consider cot 2x cot 3x + cot x cot 3x = cot (2x+x)(cot 2x+ cot x)= ((cot x cot 2x−1)/(cot 2x+cot x))(cot 2x+cot x) = cot x cot 2x−1 adding cot x cot 2x both sides, cot x cot 2x+cot 2x cot 3x + cot x cot 3x= 2cot x cot 2x−1 cot x cot 2x +cot 2x cot 3x+cot x cot 3x+2= 2cot x cot 2x+1 ∴ 2(((cos x)/(sin x)))(((cos 2x)/(sin 2x)))+1=2(((cos x)/(sin x)))(((cos^2 x−sin^2 x)/(2sin xcos x)))+1 = cot^2 x ∴ cot x cot 2x + cot 2x cot 3x = cot^2 x−cot xcot 3x cot x cot 2x+cot 2xcot 3x=cot x(cot x−cot 3x)
considercot2xcot3x+cotxcot3x=cot(2x+x)(cot2x+cotx)=cotxcot2x−1cot2x+cotx(cot2x+cotx)=cotxcot2x−1addingcotxcot2xbothsides,cotxcot2x+cot2xcot3x+cotxcot3x=2cotxcot2x−1cotxcot2x+cot2xcot3x+cotxcot3x+2=2cotxcot2x+1∴2(cosxsinx)(cos2xsin2x)+1=2(cosxsinx)(cos2x−sin2x2sinxcosx)+1=cot2x∴cotxcot2x+cot2xcot3x=cot2x−cotxcot3xcotxcot2x+cot2xcot3x=cotx(cotx−cot3x)

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