Question

please solve thise math​

Answer 1

$Given \: \: Question \: \: Is \: \\ \\ \tan(x) + \sin(x) = m \\ and \\ \tan(x) - \sin(x) = n \\ \\ multiply \: \: both \: the \: two \: equations \\ \\ mn = ( \tan(x) + \sin(x) )( \tan(x) - \sin(x) ) \\ \\ mn = ( \tan {}^{2} (x) - \sin {}^{2} (x) ) \\ becoz \: \: \: (x - y)(x + y) = (x {}^{2} - y {}^{2} ) \\ \\ mn = ( \frac{ \sin {}^{2} (x) }{ \cos {}^{2} (x) } - \sin {}^{2} (x) ) \\ \\becoz \: \: \: \: \: \: \: \: \tan(x) = \frac{ \sin(x) }{ \cos(x) } \\ \\ mn = \sin {}^{2} (x) ( \frac{1}{ \cos {}^{2} (x) } - 1) \\ \\ mn = \sin {}^{2} (x) ( \frac{1 - \cos {}^{2} (x) }{ \cos {}^{2} (x) } ) \\ \\ mn = \sin {}^{2} (x) ( \frac{ \sin {}^{2} (x) }{ \cos {}^{2} (x) } ) \\ \\ becoz \: \: \: 1 - \cos {}^{2} (x) = \sin {}^{2} (x) \\ \\ mn = \sin {}^{2} (x) \times \cos {}^{2} (x) \\ \\ becoz \: \: \: \frac{ \sin {}^{2} (x) }{ \cos {}^{2} (x) } = \tan {}^{2} (x) \\ \\ therefore \: \: \: \: mn = \sin {}^{2} (x) \times \tan {}^{2} (x)$