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prove that: tan3A.tan2A.tanA = tan3A - tan2A - tanA
tan2A can be written as tan(3A-A) now we have tan(a-b)=tana-tanb/1+tanatanb here a =2A & b=A tan2A=tan(3A-A)=tan3A-tanA/1+tan3AtanA tan2A(1+tan3Atan2A) = tan3A-tanA tan2A + tan2Atan3AtanA = tan3A -tanA or tan2Atan3AtanA = tan3A -tanA-tan2A hence proved
tan2A can be written as tan(3A-A)
now we have tan(a-b)=tana-tanb/1+tanatanb
here a =2A & b=A
tan2A=tan(3A-A)=tan3A-tanA/1+tan3AtanA
tan2A(1+tan3Atan2A) = tan3A-tanA
tan2A + tan2Atan3AtanA = tan3A -tanA or
tan2Atan3AtanA = tan3A -tanA-tan2A
hence proved
Hello student,tan2A can be written as tan(3A-A)now we have tan(a-b)=tana-tanb/1+tanatanbhere a =2A & b=Atan2A=tan(3A-A)=(tan3A-tanA)/(1+tan3AtanA)tan2A(1+tan3Atan2A) = tan3A-tanAtan2A + tan2Atan3AtanA = tan3A -tanAtan2Atan3AtanA = tan3A -tanA-tan2AHence provedThanksI hope above solution will clear your all doubts.Please feel free to post and ask as much doubts as possible.All the best.
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