Question

if Asin alpha=Bsin(alpha+2beta) then B+A/B-A is equal to.
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Answer 1

SOLUTION

GIVEN

$\sf{A \sin \alpha = B \sin( \alpha + 2 \beta )}$

TO DETERMINE

$\displaystyle \sf{ \frac{ B + A\: }{B - A} \: }$

FORMULA TO BE IMPLEMENTED

$\displaystyle \sf{1. \: \: \sin \alpha + \sin \beta = 2 \sin \bigg( \frac{ \alpha + \beta }{2} \bigg) \cos \bigg( \frac{ \alpha - \beta }{2} \bigg)\: }$

$\displaystyle \sf{2. \: \: \sin \alpha - \sin \beta = 2 \cos \bigg( \frac{ \alpha + \beta }{2} \bigg) \sin \bigg( \frac{ \alpha - \beta }{2} \bigg)\: }$

EVALUATION

$\sf{A \sin \alpha = B \sin( \alpha + 2 \beta )}$

$\displaystyle \implies \sf{ \frac{B}{ A} = \frac{\sin \alpha \: }{\sin( \alpha + 2 \beta )} }$

$\displaystyle \implies \sf{ \frac{B + A}{ B - A} = \frac{\sin \alpha +\sin( \alpha + 2 \beta ) \: }{\sin \alpha - \sin( \alpha + 2 \beta )} }$

$\displaystyle \implies \sf{ \frac{B + A}{ B - A} = \frac{2\sin( \alpha + \beta ) \cos \beta \: }{ - 2\cos( \alpha + \beta ) \sin \beta } }$

$\displaystyle \implies \sf{ \frac{B + A}{ B - A} = - \frac{\sin( \alpha + \beta ) \cos \beta \: }{ \cos( \alpha + \beta ) \sin \beta } }$

$\displaystyle \implies \sf{ \frac{B + A}{ B - A} = - {\tan( \alpha + \beta ) \cot \beta \: } }$

FINAL ANSWER

$\displaystyle \sf{ \frac{B + A}{ B - A} = - {\tan( \alpha + \beta ) \cot \beta \: } }$

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