Question

an angle a is divided into two parts b and c such that tan b : tan c = x:y ; prove that sin ( b - c) = (x+y/x-y) sin a​

Answer 1

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• An angle a is divided into two parts b and c such that tan b : tan c = x:y ; prove that sin ( b + c ) = ( x+y/x-y ) sin ( b - c )

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$\\ \dashrightarrow \: \: \sf \tan(b) : \tan(c) = x : y$

$\sf \dashrightarrow \frac{ \tan(b) }{ \tan(c) } = \frac{x}{y}$

$\sf \dashrightarrow \frac{ \sin(b) \cos(c) }{ \sin(c) \cos(b) } = \frac{x}{y}$

$\sf \dashrightarrow \frac{ \sin(b) \cos(c) + \cos(b) \sin(c) }{ \sin(b) \cos(c) - \cos(b) \sin(c) } = \frac{x + y}{x - y}$

$\sf \dashrightarrow \frac{ \sin(b + c) }{ \sin(b - c) } = \frac{x + y}{x - y}$

$\sf \dashrightarrow \: \blue{\sin(b + c) = \bigg( \frac{x + y}{x - y} \bigg) \sin(b - c)} \: \: \: \: \: \: \: \: \: \bf{Proved.}$

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