Question #5df8b

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Question #5df8b

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Answer 1 · 2016-12-21T15:06:08

GIVEN

$\frac{tan (A + B)}{tan (A - B)} = λ$ $tan(A+B)/tan(A-B) = lamda$

$\Rightarrow \frac{sin (A + B) cos (A - B)}{sin (A - B) cos (A + B)} = λ$ $=>(sin(A+B)cos(A-B))/(sin(A-B)cos(A+B)) = lamda$

By dividendo and componendo

$\Rightarrow \frac{(sin (A + B) cos (A - B)) - (sin (A - B) cos (A + B))}{(sin (A + B) cos (A - B)) + (sin (A - B) cos (A + B))} = \frac{λ - 1}{λ + 1}$ $=>((sin(A+B)cos(A-B))-(sin(A-B)cos(A+B)))/((sin(A+B)cos(A-B))+(sin(A-B)cos(A+B))) = (lamda-1)/(lambda+1)$

$\Rightarrow \frac{sin (A + B - A + B)}{sin (A + B + A - B)} = \frac{λ - 1}{λ + 1}$ $=>(sin(A+B-A+B))/(sin(A+B+A-B))=(lambda-1)/(lambda+1)$

$\Rightarrow \frac{sin (2 B)}{sin (2 A)} = \frac{λ - 1}{λ + 1}$ $=>(sin(2B))/(sin(2A))=(lambda-1)/(lambda+1)$

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Question #5df8b

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