Question

If neither A nor A+B is an odd multiple of π÷2 and if m sin B = n sin ( 2A+ B) , then prove that (m+n) tan A=(m - n) tan ( A+B)​

Answer 1

We have:

m.sin(B)=n.sin(2A+B)

⇒m/n=sin(2A+B)/sin(B)

Applying components and dividing, we get:

m+n/m−n=sin(2A+B)+sin(B)/sin(2A+B)−sin(B)

⇒m+n/m−n=2.sin(2A+B+B/2).cos(2A+B−B/2)/2.cos(2A+B+B/2).sin(2A+B−B/2)

⇒m+n/m−n=sin(2A+2B/2).cos(2A/2)/cos(2A+2B/2).sin(2A.2)

⇒m+n/m−n=sin(A+B).cos(A)/cos(A+B).sin(A)

⇒m+nm−n=tan(A+B)/tan(A)

y⇒(m+n) tan(A)=(m−n) tan(A+B)

Answer 2

Answer:

Dear Student,

Please find below the solution to the asked query:

We have:m.sin(B)=n.sin(2A+B)⇒mn=sin(2A+B)sin(B)Applying componando and dividendo, we get:m+nm−n=sin(2A+B)+sin(B)sin(2A+B)−sin(B)⇒m+nm−n=2.sin(2A+B+B2).cos(2A+B−B2)2.cos(2A+B+B2).sin(2A+B−B2)⇒m+nm−n=sin(2A+2B2).cos(2A2)cos(2A+2B2).sin(2A2)⇒m+nm−n=sin(A+B).cos(A)cos