If $$m = {\text{cosec}}\,\theta - \sin \theta $$     and $$n = \sec \theta - \cos \theta ,$$    then $${m^{\frac{2}{3}}} + {n^{\frac{2}{3}}} = $$

Solution :
We have,
$$\eqalign{ & mn\, = \left( {{\text{cosec}}\,\theta - \sin \,\theta } \right)\left( {\sec \,\theta - \cos \,\theta } \right) \cr & = \,\left( {\frac{1}{{\sin \,\theta }} - \sin \,\theta } \right)\left( {\frac{1}{{\cos \,\theta }} - \cos \,\theta } \right) \cr & = \,\frac{{1 - {{\sin }^2}\theta }}{{\sin \,\theta }} \times \frac{{1 - {{\cos }^2}\theta }}{{\cos \,\theta }} \cr & = \,\frac{{{{\cos }^2}\theta }}{{\sin \,\theta }} \times \frac{{{{\sin }^2}\theta }}{{\cos \,\theta }} = \sin \,\theta \cdot \cos \,\theta \cr & \therefore {m^{\frac{2}{3}}} + {n^{\frac{2}{3}}} = {\left( {\frac{{{{\cos }^2}\theta }}{{\sin \,\theta }}} \right)^{\frac{2}{3}}} + {\left( {\frac{{{{\sin }^2}\theta }}{{\cos \,\theta }}} \right)^{\frac{2}{3}}} \cr & = \frac{{{{\cos }^{\frac{4}{3}}}\theta }}{{{{\sin }^{\frac{2}{3}}}\theta }} + \frac{{{{\sin }^{\frac{4}{3}}}\theta }}{{{{\cos }^{\frac{2}{3}}}\theta }} = \frac{{{{\cos }^2}\theta + {{\sin }^2}\theta }}{{{{\left( {\sin \,\theta \cdot \cos \,\theta } \right)}^{\frac{2}{3}}}}} \cr & = \,\frac{1}{{{{\left( {mn} \right)}^{\frac{2}{3}}}}} = {\left( {mn} \right)^{ - \frac{2}{3}}} \cr} $$