Question

If tan(A+B) =3tanA

then show that sin(2A +2B)+sin2A=2sin²B

Answer 1

Given : (A+B)=3tanA
sin(A+B)cosA=3sinAcos(A+B)

expand :(sinAcosB+cosAsinB)*cosA=3sinA*(cosAcosB-sinAsinB)
= (cosA)^2*(sinB)=2sinAcosAcosB-3(sinA)^2*sinB
= sinB*(1+2(sinA)^2)=sin2AcosB
= 2sin2B-sin2A=sin2Acos2B+cos2Asin2B
= sin2B*(2-cos2A)=sin2A*(1+cos2B)
= 2sinBcosB*(1+2(sinA)^2)=sin2A*(cosB)^2
=2sin2b-sin2a=sin2(a+b)
= 4sinbcosb-2sinacosa=2sin（a+b）cos（a+b）
=2sinbcosb-sinacosa=（sinacosb+cosasinb）*（cosacosb-sinasinb）
=2sinbcosb-sinacosa=sinacosa(cosb)^2-sinbcosb(sina)^2
+sinbcosb(cosa)^2-sinacosa(sinb)^2
=sinbcosb(2+(sina)^2-(cosa)^2)=sinacosa(1+(cosb)^2-(sinb)^2)
=sinbcosb(3(sina)^2+(cosa)^2)=sinacosa*2(cosb)^2

The a and b are placed at both ends of the equation :-

sinbcosb/(cosb)^2=2sinacosa/3(sina)^2+(cosa)^2

Simplification, with the right of the numerator and denominator divided by (cosa) ^ 2
tanb=2tana/3(tana)^2+1
tanb*(3(tana)^2+1)=2tana
3(tana)^2*tanb+tanb=2tana
tanb=2tana-3(tana)^2*tanb

both sides of the equation tana
tana+tanb=3tana-3(tana)^2*tanb=3tana(1-tanatanb)
3tana=(tana+tanb)/(1-tanatanb)=tan(a+b)