prove

tan3A.tan2A.tanA = tan3A - tan2A - tanA

Dear student,

tan(3A) = tan(2A+A)
tan(3A) = [tan(2A) + tan(A)] / [1 - tan(A)*tan(2A)]

cross multiply
tan(3A) * [1 - tan(A)*tan(2A)] = [tan(2A) + tan(A)]
tan(3A) - tan(A)*tan(2A)*tan(3A) = tan(2A) + tan(A)
tan(A)*tan(2A)*tan(3A) = tan(3A) - tan(2A) - tan(A)

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Askiitians Expert

Sagar Singh

B.Tech, IIT Delhi

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