Question

Hey guys solve this.......​

Answer 1

Solution:-

$x = \frac{ { sin}^{3}</strong><strong>alpha</strong><strong> }{ {cos}^{2} \alpha } ...................i) \\ y = \frac{ {cos}^{3} \alpha }{ {sin}^{2} \alpha } ...................ii) \\ add \: both \: equation \\ x + y = \frac{ {sin}^{3} \alpha }{ {cos}^{2} \alpha } + \frac{ {cos}^{3} \alpha }{ {sin}^{2} \alpha } \\ x + y = \frac{ {sin}^{3} \alpha . {sin}^{2} \alpha + {cos}^{3} \alpha . {cos}^{2} \alpha }{ {( \cos \alpha sin \alpha )}^{2} } \\ now \: convert \: {sin}^{2} \alpha \: in \: 1 - {cos}^{2} \alpha and \: {cos}^{2} \alpha \\ in 1 - {sin}^{2} \alpha \\ x + y = \frac{ {sin}^{3} \alpha (1 - {cos}^{2} \alpha ) + {cos}^{3} \alpha (1 - {sin}^{2} \alpha ) }{ {cos}^{2} \alpha {sin}^{2} \alpha } \\ x + y = \frac{ {sin}^{3} \alpha - {sin}^{3} \alpha {cos}^{2} \alpha + {cos}^{3} \alpha - {cos}^{3} \alpha {sin}^{2} \alpha }{ {sin}^{2} \alpha {cos}^{2} \alpha } \\ x + y = \frac{ {sin}^{3} \alpha + {cos}^{3} \alpha - {sin}^{2} \alpha {cos}^{2} \alpha (sin \alpha + cos \alpha ) }{ {sin}^{2} \alpha {cos}^{2} \alpha } \\ x + y = \frac{(sin \alpha + cos \alpha )( {sin}^{2} \alpha + {cos}^{2} \alpha - sin \alpha cos \alpha ) - {sin}^{2} \alpha {cos}^{2} \alpha (sin \alpha + cos \alpha ) }{ {sin}^{2} \alpha {cos}^{2} \alpha } \\ x + y = \frac{(sin \alpha + cos \alpha )(1 - sin \alpha \cos \alpha ) - {sin}^{2} \alpha {cos}^{2} \alpha (sin \alpha + cos \alpha ) }{ {sin}^{2} \alpha {cos}^{2} \alpha }........i) \\ {(sin \alpha + cos \alpha )}^{2} = \frac{1}{4} \\ {sin}^{2} \alpha + {cos}^{2} \alpha + 2sin \alpha cos \alpha = \frac{1}{4} \\ 2sin \alpha cos \alpha = \frac{1}{4} - 1 \\ sin \alpha cos \alpha = \frac{ - 3}{8} \\ now \: put \: these \: values \: in \: eq.i) \\ x + y = \frac{ \frac{1}{2} (1 + \frac{3}{8}) - \frac{9}{64} \times \frac{1}{2} }{ \frac{9}{64} } \\ x + y = \frac{79}{18}$