Question

tan(A+B)=ntan (A-B) then prove that (n+1)sin2B= (n-1)sin2A

Answer 1

tan (A+B)=n tan(A-B)

sin(A+B)/cos (A+B)=nsin (A-B)/cos (A-B)

sin (A+B).cos (A-B)/sin (A-B).cos (A+B)=n
now use componendo and devedendo rule
{sin(A+B).cos (A-B)+sin (A-B).cos (A+B)}/{sin(A+B).cos (A-B)-sin (A-B).cos (A+B)}=
(n+1)/(n-1)

(sin2A+sin2B+sin2A-sin2B)/(sin2A+sin2B-sin2A+sin2B)=(n+1)/(n-1)
sin2A/sin2B=(n+1)/(n-1)
(n+1) sin2B=(n-1) sin2A

Answer 2

Answer:

hope it's helpful to u