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If sin2A= lambda sin 2B prove that (tan...

If ` sin2A= lambda sin 2B` prove that `(tan(A+B)/tan(A-B))=(lambda+1)/(lambda-1)`

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Given `sin2A=lambdasin 2B`
`rArr(Sin2A)/(sin2B)=(lambda)/(1)`
Applying componendo `&` 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))/(an(A-B))=(lambda+1)/(lambda-1)`
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