If sin A = n sin B, then frac{n - 1}{n + 1} t | vedclass

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(B) Given $\sin A = n \sin B,$ we have $\frac{n}{1} = \frac{\sin A}{\sin B}.$Applying the componendo and dividendo rule,we get $\frac{n - 1}{n + 1} =

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$	an \frac{A - B}{2}$

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(B) Given $\sin A = n \sin B,$ we have $\frac{n}{1} = \frac{\sin A}{\sin B}.$Applying the componendo and dividendo rule,we get $\frac{n - 1}{n + 1} = \frac{\sin A - \sin B}{\sin A + \sin B}.$Using the sum-to-product formulas $\sin A - \sin B = 2 \cos \frac{A + B}{2} \sin \frac{A - B}{2}$ and $\sin A + \sin B = 2 \sin \frac{A + B}{2} \cos \frac{A - B}{2},$$\frac{n - 1}{n + 1} = \frac{2 \cos \frac{A + B}{2} \sin \frac{A - B}{2}}{2 \sin \frac{A + B}{2} \cos \frac{A - B}{2}} = \cot \frac{A + B}{2} 	an \frac{A - B}{2}.$Multiplying both sides by $	an \frac{A + B}{2},$ we get $\frac{n - 1}{n + 1} 	an \frac{A + B}{2} = 	an \frac{A - B}{2} \cot \frac{A + B}{2} 	an \frac{A + B}{2} = 	an \frac{A - B}{2}.$

If $\sin A = n \sin B,$ then $\frac{n - 1}{n + 1} \tan \frac{A + B}{2} = $

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