Question

Help Help anyone solve this

Answer 1

let's replace A by theta and alpha by B => SinA=nSin(A+2B) => SinA/Sin(A+2B) =n => 1/n =Sin(A+2B)/SinA => Apply compendo and dividendo
we get SinA+Sin(A+2B)/SinA-Sin(A+2B) =1+n/1-n =>
2Sin(A+B)Cos(-B)/2Cos(A+B)Sin(B) =1+n/1-n =>
Tan(A+B)/Tan(B) =1+n/1-n => Tan(A+B) =1+n/1-n (Tan B)
=> Tan(theta+alpha) =1+n/1-n(Tan alpha) So (D) is the correct option.

Answer 2

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