1/sec-tan-1/cos=1/cos-1sec+tan

According to the question,
L.H.S. = {tex}\frac { 1 } { ( \sec \theta - \tan \theta ) } - \frac { 1 } { \cos \theta }{/tex}
{tex}= \frac { 1 } { ( \sec \theta - \tan \theta ) } \times \frac { ( \sec \theta + \tan \theta ) } { ( \sec \theta + \tan \theta ) } - \sec \theta{/tex}
{tex}= \frac { ( \sec \theta + \tan \theta ) } { \left( \sec ^ { 2 } \theta - \tan ^ { 2 } \theta \right) } - \sec \theta{/tex}
= (sec{tex}\theta{/tex} + tan{tex}\theta{/tex}) - sec{tex}\theta{/tex} [{tex}\therefore{/tex} {tex}sec^2\theta - tan^2\theta = 1{/tex}]
= tan{tex}\theta{/tex}.
R.H.S. = {tex}\frac { 1 } { \cos \theta } - \frac { 1 } { ( \sec \theta + \tan \theta ) }{/tex}
{tex}= \sec \theta - \frac { 1 } { ( \sec \theta + \tan \theta ) } \times \frac { ( \sec \theta - \tan \theta ) } { ( \sec \theta - \tan \theta ) }{/tex}
= sec{tex}\theta{/tex} - (sec{tex}\theta{/tex} - tan{tex}\theta{/tex}) [{tex}\therefore{/tex}{tex} sec^2\theta - tan^2\theta = 1{/tex}]
= tan{tex}\theta{/tex}.
{tex}L.H.S. = R.H.S.{/tex}