Respuesta :

Solution,

As we have,

⇒2SinACosB=Sin(A+B)+Sin(A-B)

Now,

[tex]2sin7\alpha .cos2\alpha =sin9\alpha +sin5\alpha \\\\LHS=2sin7\alpha .cos2\alpha \\\\LHS=sin(7\alpha +2\alpha )+sin (7\alpha -2\alpha )\\\\LHS=sin9\alpha +sin5\alpha[/tex]

Hence, LHS=RHS