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Question
Prove the following:
If sin 2A = λsin 2B then prove that `(tan("A" + "B"))/(tan("A" - "B")) = (lambda + 1)/(lambda - 1)`
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Solution
sin 2A = λsin 2B
∴ `(sin 2"A")/(sin 2"B") = lambda/1`
∴ `(sin2"A" + sin2"B")/(sin2"A" - sin2"B") = (lambda + 1)/(lambda - 1)`
∴ `(2sin((2"A" + 2"B")/2)*cos((2"A" - 2"B")/2))/(2cos ((2"A" + 2"B")/2)*sin((2"A" - 2"B")/2)) = (lambda + 1)/(lambda - 1)`
∴ `(sin("A" + "B")*cos("A" - "B"))/(cos("A" + "B")*sin("A" - "B")) = (lambda + 1)/(lambda - 1)`
∴ tan(A + B) · cot(A – B) = `(lambda + 1)/(lambda - 1)`
∴ `(tan("A" + "B"))/(tan("A" - "B")) = (lambda + 1)/(lambda - 1)`.
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