Hello student, 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 tan2Atan3AtanA = tan3A -tanA-tan2A Hence proved Thanks I 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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