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Question

Prove that following identities:

tan A+tan(60+A)tan(60A)=3 tan 3A

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Solution

tan A+tan(60+A)tan(60A)=3 tan 3A

LHS=tan A+tan(60+A)tan(60A)

=tan A+tan 60+tan A1tan 60 tan Atan 60tan A1+tan 60 tan A=tan A+3+tan A13tan A3tan A1+3tan A=tan A+[3+3 tan A+tan A+3 tan2 A+3+3 tan A+tan A3 tan2 A(13 tan A)(1+3 tan A)]=tan A+8 tan A13 tan2 A=tan A3 tan3 A+8 tan A13 tan2 A=9 tan A3 tan3 A13 tan2 A=3(3 tan Atan3 A13 tan2 A)=3 tan 3A

So,

tan A+tan (60+A)tan (60A)=3 tan 3A


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