3sin B=Sin(2A+B) prove that: i) tan(A+B)=2tanA ii) 2sin B= cos A sin(A+B) iii)sin B= sin A cos(A+B) iv) [cot A + cot(A+B)] [cot B -3 cot(2A+B)] =6 Share with your friends Share 2 Ghanshyam Dhakar answered this Dear student, i. 3sinB = sin2A+B⇒3sinA+B-A = sinA+B+A use sinA±B=sinAcosB±cosAsinB 3sinA+BcosA-cosA+BsinA = sinA+B cosA + cosA+BsinA ⇒2sinA+BcosA=4cosA+BsinA ⇒sinA+BcosA+B=2×sinAcosA ⇒tanA+B =2tanA .....1ii.3sinB = sin2A+B⇒4sinB-sinA+B-A= sinA+B+A ⇒4sinB - sinA+BcosA+cosA+BsinA = sinA+BcosA+cosA+BsinA ⇒4sinB = 2sinA+BcosA⇒2sinB = sinA+BcosA .....2iii. 3sinB = sin2A+B⇒2sinB+sinA+B-A = sinA+B+A⇒2sinB + sinA+BcosA-cosA+BsinA = sinA+BcosA+cosA+BsinA ⇒2sinB = 2cosA+BsinA⇒sinB = cosA+BsinAiv.LHS=cotA+cotA+BcotB-3cot2A+B=cotA+cotA+BcotB-3cotA+B+Ause cotx+y = cotx coty-1cotx+cotyLHS=cotA+cotA+BcotB-3cotA+BcotA-1cotA+B+cotA=cotA+cotA+BcotBcotA+B+cotB cotA-3cotA+BcotA+3cotA+B+cotAuse eq 1tanA+B =2tanA ⇒2cotA+B =cotALHS=cotBcotA+B+2cotB cotA+B-3cotA+BcotA+3=3cotBcotA+B-3cotA+BcotA+3=3cotA+BcotB-cotA+3=32cotAcosBsinB-cosAsinA+3=32cosAsinAsinAcosB-sinBcosAsinAsinB+3=32cosAsinAsinAcosB+sinBcosA-2sinBcosAsinAsinB+3=32cosAsinAsinA+B-2sinBcosAsinAsinB+3=321sinAcosAsinA+B-2sinBcos2AsinAsinB+3use eq2LHS =321sinA2sinB-2sinBcos2AsinAsinB+3=321sinA2sinB1-cos2AsinAsinB+3=321sinA2sinBsin2AsinAsinB+3=3+3=6 =RHS Hope this information will clear your doubts. If you have any more doubts just ask here on the forum and our experts will try to help out as soon as possible. Regards -1 View Full Answer
