If `sin 2A=lambda sin 2B`, prove that : `(tan (A+B))/(tan(A-B)) = (lambda +1)/(lambda-1)`

Explanation

Given `sin2A = lambda sin2B`
`rArr (sin2A)/(sin2B) = lambda/1`
Applying componendo and dividendo,
`(sin2A+sin2B)/(sin2B-sin2A) = (lambda+1)/(1-lambda)`
`rArr ((2sin(2A+2B)/(2))cos(2A-2B)/(2))/(2cos(2B+2A)/(2)sin(2B-2A)/(2)) = (lambda+1)/(1-lambda)`
`rArr (sin(A+B)cos(A-B))/(cos(A+B)sin(-(A-B)))=(lambda+1)/(1-lambda) rArr (sin(A+B)cos(A-B))/(cos(A+B)x-sin(A-B)) = (lambda+1)/(-(lambda-1))`
`rArr (sin(A+B)cos(A-B))(cos(A+B)sin(A-B))=(lambda+1)/(lambda-1) rArr tan(A+B) cot(A-B)=(lambda+1)/(lambda-1)`
`rArr (tan(A+B))/(tan(A-B)) = (lambda+1)/(lambda-1)`