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If $\sin B=\large\frac{1}{5}$$\sin(2A+B)$ then $\large\frac{\tan(A+B)}{\tan A}$ is equal to

$(a)\;\large\frac{5}{3}$$\qquad(b)\;\large\frac{2}{3}$$\qquad(c)\;\large\frac{3}{2}$$\qquad(d)\;\large\frac{3}{5}$

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1 Answer

$\sin B=\large\frac{1}{5}$$\sin(2A+B)$
$\Rightarrow \large\frac{\sin(2A+B)}{\sin B}$$=5$
$\Rightarrow \large\frac{\sin(2A+B)+\sin B}{\sin(2A+B)-\sin B}$$=\large\frac{6}{4}$
$\Rightarrow \large\frac{\sin(A+B)\cos A}{\cos(A+B)\sin A}=\frac{3}{2}$
$\Rightarrow \large\frac{\tan(A+B)}{\tan A}=\frac{3}{2}$
Hence (c) is the right option.
answered Oct 8, 2013 by sreemathi.v
 

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