Question

tan(a+b)=3tana show thatsin (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) 

hence it is proved