Question

if (sin theta)/x=(cos theta)/y then (sin theta) - (cos theta)=?

Answer 1

$\frac{ \sin( \alpha ) }{x} = \frac{ \cos( \alpha ) }{y} \\ take \: square \: on \: both \: side \\ \\ \frac{ sin^{2}{ \alpha } }{ {x}^{2} } = \frac{cos^{2}{ \alpha } }{ {y}^{2} } \\ \\ we \: know \: that \\ \: sin^{2} \alpha = 1 - cos^{2} \alpha \\ \\ put \: the \: value \: in \: upper \: equation \\ \\ 1 -cos^{2} \alpha = {x}^{2} \frac{cos^{2} \alpha }{ {y}^{2} } \\ \\ adust \: the \: cos^{2} \alpha \\ \\ 1 = cos^{2} \alpha ( 1 + \frac{ {x}^{2} }{ {y}^{2} } ) \\ \\ 1 = cos^{2} \alpha ( \frac{ {x}^{2} + {y}^{2} }{ {y}^{2} } ) \\ \\ cos^{2} \alpha = \frac{ {y}^{2} }{ {x}^{2} + {y}^{2} } .......eq(1) \\ \\ cos \alpha = \frac{y}{ \sqrt{ {x}^{2} + {y}^{2} } } .....eq(2) \\ \\ from \: eq(1) \: we \: get \\ sin^{2} \alpha = \frac{ {x}^{2} }{ {x}^{2} + {y}^{2} } \\ \\ sin \: \alpha = \frac{x}{ \sqrt{ {x}^{2} + {y}^{2} } } .........eq(3) \\ \\ subtract \: eq(2) \: from \: eq(3) \\ \\ \sin( \alpha ) - \cos( \alpha ) = \frac{x - y}{ \sqrt{ {x}^{2} + {y}^{2} } }$