Question

. Prove that sino + cose + 3sin’ocos'o = 1.​

Answer 1

Step-by-step explanation:

sinB=3sin(2A+B)

⇒sin(A+B−A)=3sin(A+A+B)

⇒sin(A+B)cosA−cos(A+B)sin(B)=3[sin(A+B)cosA+cos(A+B)sin(B)]

divide both side by cos(A+B).cosA

⇒tan(A+B)−tan(A)=3tan(A+B)+3tan(A)

⇒4tan(A)+2tan(A+B)=0

∴2tan(A)+tan(A+B)=0