If b sin alpha = a sin (alpha + 2beta),then f | vedclass

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(C) Given $b \sin \alpha = a \sin (\alpha + 2\beta)$.Dividing both sides,we get $\frac{a}{b} = \frac{\sin \alpha}{\sin (\alpha + 2\beta)}$.Using the c

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$\frac{-\cot \beta}{\cot (\alpha + \beta)}$

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(C) Given $b \sin \alpha = a \sin (\alpha + 2\beta)$.Dividing both sides,we get $\frac{a}{b} = \frac{\sin \alpha}{\sin (\alpha + 2\beta)}$.Using the componendo and dividendo rule,$\frac{a + b}{a - b} = \frac{\sin \alpha + \sin (\alpha + 2\beta)}{\sin \alpha - \sin (\alpha + 2\beta)}$.Applying the sum-to-product formulas $\sin C + \sin D = 2 \sin \frac{C+D}{2} \cos \frac{C-D}{2}$ and $\sin C - \sin D = 2 \cos \frac{C+D}{2} \sin \frac{C-D}{2}$:$\frac{a + b}{a - b} = \frac{2 \sin (\alpha + \beta) \cos (-\beta)}{2 \cos (\alpha + \beta) \sin (-\beta)}$.Since $\cos(-\beta) = \cos \beta$ and $\sin(-\beta) = -\sin \beta$:$\frac{a + b}{a - b} = \frac{2 \sin (\alpha + \beta) \cos \beta}{-2 \cos (\alpha + \beta) \sin \beta} = -	an (\alpha + \beta) \cot \beta$.This can be rewritten as $-\frac{\cot \beta}{\cot (\alpha + \beta)}$.

If $b \sin \alpha = a \sin (\alpha + 2\beta)$,then $\frac{a + b}{a - b} = $

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