Prove that tan theta + sec …

According to the question,

L.H.S. = {tex}\frac { \sec \theta + \tan \theta - 1 } { \tan \theta - \sec \theta + 1 }{/tex}
{tex}= \frac { ( \sec \theta + \tan \theta ) - \left( \sec ^ { 2 } \theta - \tan ^ { 2 } \theta \right) } { ( \tan \theta - \sec \theta + 1 ) }{/tex} [{tex}\because{/tex} sec²{tex}\theta{/tex} - tan²{tex}\theta{/tex} = 1]
{tex}= \frac { ( \sec \theta + \tan \theta ) [ 1 - ( \sec \theta - \tan \theta ) ] } { ( \tan \theta - \sec \theta + 1 ) }{/tex}
{tex}= \frac { ( \sec \theta + \tan \theta ) ( \tan \theta - \sec \theta + 1 ) } { ( \tan \theta - \sec \theta + 1 ) } = ( \sec \theta + \tan \theta ){/tex}
{tex}= \left( \frac { 1 } { \cos \theta } + \frac { \sin \theta } { \cos \theta } \right) = \frac { ( 1 + \sin \theta ) } { \cos \theta } = \frac { ( 1 + \sin \theta ) } { \cos \theta } \times \frac { ( 1 - \sin \theta ) } { ( 1 - \sin \theta ) }{/tex}
{tex}= \frac { \left( 1 - \sin ^ { 2 } \theta \right) } { \cos \theta ( 1 - \sin \theta ) } = \frac { \cos ^ { 2 } \theta } { \cos \theta ( 1 - \sin \theta ) } = \frac { \cos \theta } { ( 1 - \sin \theta ) }{/tex} = R.H.S.
{tex}\therefore{/tex} {tex}\frac { \sec \theta + \tan \theta - 1 } { \tan \theta - \sec \theta + 1 } = \frac { \cos \theta } { ( 1 - \sin \theta ) }{/tex}