Prove that. tan/1-cot+cot/1-tan=(1+sec cosec)

LHS = {tex}\frac { \tan \theta } { 1 - \cot \theta } + \frac { \cot \theta } { 1 - \tan \theta }{/tex}
{tex}= \frac { \frac { \sin \theta } { \cos \theta } } { 1 - \frac { \cos \theta } { \sin \theta } } + \frac { \frac { \cos \theta } { \sin \theta } } { 1 - \frac { \sin \theta } { \cos \theta } }{/tex} {tex}\left[ \because \tan \theta = \frac { \sin \theta } { \cos \theta } , \cot \theta = \frac { \cos \theta } { \sin \theta } \right]{/tex}
{tex}= \frac { \sin ^ { 2 } \theta } { \cos \theta ( \sin \theta - \cos \theta ) } + \frac { \cos ^ { 2 } \theta } { \sin \theta ( \cos \theta - \sin \theta ) }{/tex}
{tex}= \frac { \sin ^ { 2 } \theta } { \cos \theta ( \sin \theta - \cos \theta ) } - \frac { \cos ^ { 2 } \theta } { \sin \theta ( \sin \theta - \cos \theta ) }{/tex}
{tex}= \frac { \sin ^ { 3 } \theta - \cos ^ { 3 } \theta } { \sin \theta \cos \theta ( \sin \theta - \cos \theta ) }{/tex} 
{tex}= \frac { ( \sin \theta - \cos \theta ) \left( \sin ^ { 2 } \theta + \cos ^ { 2 } \theta + \sin \theta \cos \theta \right) } { ( \sin \theta - \cos \theta ) \sin \theta \cos \theta }{/tex} [ a³ - b³ = (a - b)(a² + ab + b²) ]
{tex}= \frac { 1 + \sin \theta \cos \theta } { \sin \theta \cos \theta }{/tex}
{tex}= \frac { 1 } { \sin \theta \cos \theta } + 1 = 1 + \sec \theta cosec \theta{/tex} = RHS
therefore, RHS = LHS