If sinx+3a=3sina-x, then
tanx=tana
tanx=tan2a
tanx=tan3a
tanx=3tana
Explanation for the correct option
Here use the identity sin(A+B)=sin(A)cos(B)+sin(B)cos(A) in the Left-hand side (LHS )and identity sin(A-B)=sin(A)cos(B)-sin(B)cos(A) in the RHS.
β΄sinxcos3a+sin3acosx=3sinacosx-sinxcosaβsinxcos3a+3sinxcosa=3sinacosx-sin3acosxβsinxcos3a+3cosa=cosx3sina-sin3a
We know that cos(3A)=-3cos(A)+4cos3(A), we have
cos(3A)+3cos(A)=4cos3(A) and sin(3A)=3sin(A)-4sin3(A), then
4sin3(A)=3sin(A)-sin(3A)
Now, the equation sinxcos3a+3cosa=cosx3sina-sin3a can be written as
β΄sinx4cos3a=cosx4sin3aβtanx=tan3a
Hence, the correct option is (C).