$\begin{array}{l} \dfrac{{{{\sin }^2}x}}{{\sin x - \cos x}} + \dfrac{{\sin x - \cos x}}{{1 - {{\tan }^2}x}}\\ = \dfrac{{{{\sin }^2}x - {{\sin }^2}x.{{\tan }^2}x + \left( {\sin x - \cos x} \right)\left( {\sin x + \cos x} \right)}}{{\left( {\sin x - \cos x} \right)\left( {1 - {{\tan }^2}x} \right)}}\\ = \dfrac{{{{\sin }^2}x - {{\sin }^2}x.{{\tan }^2}x + {{\sin }^2}x - {{\cos }^2}x}}{{\left( {\sin x - \cos x} \right)\left( {1 - {{\tan }^2}x} \right)}} = \dfrac{{{{\tan }^2}x.{{\cos }^2}x - {{\sin }^2}x.{{\tan }^2}x + {{\sin }^2}x - {{\cos }^2}x}}{{\left( {\sin x - \cos x} \right)\left( {1 - {{\tan }^2}x} \right)}}\\ = \dfrac{{{{\tan }^2}x\left( {{{\cos }^2}x - {{\sin }^2}x} \right) + {{\sin }^2}x - {{\cos }^2}x}}{{\left( {\sin x - \cos x} \right)\left( {1 - {{\tan }^2}x} \right)}} = \dfrac{{\left( {{{\sin }^2}x - {{\cos }^2}x} \right)\left( {1 - {{\tan }^2}x} \right)}}{{\left( {\sin x - \cos x} \right)\left( {1 - {{\tan }^2}x} \right)}}\\ = \dfrac{{\left( {\sin x - \cos x} \right)\left( {\sin x + \cos x} \right)\left( {1 - {{\tan }^2}x} \right)}}{{\left( {\sin x - \cos x} \right)\left( {1 - {{\tan }^2}x} \right)}} = \sin x + \cos x \end{array}$
