Prove cot^2 theta (sec theta -1)/1+sin …

LHS = {tex}\frac { \cot ^ { 2 } \theta ( \sec \theta - 1 ) } { ( 1 + \sin \theta ) } + \frac { \sec ^ { 2 } \theta ( \sin \theta - 1 ) } { ( 1 + \sec \theta ) }{/tex}
{tex}= \frac { \cot ^ { 2 } \theta ( \sec \theta - 1 ) ( 1 + \sec \theta ) + \sec ^ { 2 } \theta ( \sin \theta - 1 ) ( 1 + \sin \theta ) } { ( 1 + \sin \theta ) ( 1 + \sec \theta ) }{/tex}
{tex}= \frac { \cot ^ { 2 } \theta \left( \sec ^ { 2 } \theta - 1 \right) + \sec ^ { 2 } \theta \left( \sin ^ { 2 } \theta - 1 \right) } { ( 1 + \sin \theta ) ( 1 + \sec \theta ) }{/tex}
{tex}= \frac { \cot ^ { 2 } \theta \tan ^ { 2 } \theta + \sec ^ { 2 } \theta \left( - \cos ^ { 2 } \theta \right) } { ( 1 + \sin \theta ) ( 1 + \sec \theta ) }{/tex}
{tex}= \frac { \cot ^ { 2 } \theta \tan ^ { 2 } \theta - \sec ^ { 2 } \theta \cos ^ { 2 } \theta } { ( 1 + \sin \theta ) ( 1 + \sec \theta ) }{/tex}
{tex}= \frac { \cot ^ { 2 } \theta \times \frac { 1 } { \cot ^ { 2 } \theta } - \sec ^ { 2 } \theta \times \frac { 1 } { \sec ^ { 2 } \theta } } { ( 1 + \sin \theta ) ( 1 + \sec \theta ) }{/tex}
{tex}= \frac { 1 - 1 } { ( 1 + \sin \theta ) ( 1 + \sec \theta ) }{/tex}
= 0
= RHS