If in a triangle ABC , a tanA +btanB =(a+b) tan(A+B) then,
1. A=B=C 2. C=A 3.A=B 4.B=C 2

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If in a triangle ABC , a tanA +btanB =(a+b) tan(A+B)
then,
1 A=B=C 2 C=A 3 A=B 4 B=C 2

Neha Sethi · Accepted Answer

Dear student
atanA+btanB=(a+b)tanA+B2atanA+btanB=atanA+B2+btanA+B2atanA-tanA+B2=btanA+B2-tanBasinAcosA-sinA+B2cosA+B2=bsinA+B2cosA+B2-sinBcosBasinAcosA+B2-sinA+B2cosAcosAcosA+B2=bsinA+B2 cosB-sinB cosA+B2cosB cosA+B2asinA-A+B2cosA=bsinA+B2-BcosBasinA-B2cosA=bsinA-B2cosBacosB=bcosANow, since a and b are some constants
So, Let a=b=kSo, we getcosB=cosAB=ANote:sinx-y=sinxcosy-cosxsiny
Regards

If in a triangle ABC , a tanA +btanB =(a+b) tan(A+B) then, 1. A=B=C 2. C=A 3.A=B 4.B=C 2

If in a triangle ABC , a tanA +btanB =(a+b) tan(A+B) then,
1. A=B=C 2. C=A 3.A=B 4.B=C 2