Proove that:- tan÷1-cot+cot÷1-tan=1+tan+cot

{tex}\frac { \tan \theta } { 1 - \cot \theta } + \frac { \cot \theta } { 1 - \tan \theta } = 1 + \tan \theta + \cot \theta{/tex}
{tex}\text { L.H.S. } = \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}= \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}{tex}\left[ {\because {a^3} - {b^3} = (a - b)({a^2} + ab + {b^2}} \right){/tex}
{tex}= \frac { \sin ^ { 2 } \theta } { \sin \theta \cos \theta } + \frac { \cos ^ { 2 } \theta } { \sin \theta \cos \theta } + \frac { \sin \theta \cos \theta } { \sin \theta \cos \theta }{/tex}
{tex}= \tan \theta + \cot \theta + 1 = 1 + \tan \theta + \cot \theta = R . H S \text { proved }{/tex}
Since, {tex}\tan A = \frac{{\sin A}}{{\cos A}}{/tex}

{tex}\cot A = \frac{{\cos A}}{{\sin A}}{/tex}