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

We have to prove that {tex} \frac { 1 + \sec \theta - \tan \theta } { 1 + \sec \theta + \tan \theta } = \frac { 1 - \sin \theta } { \cos \theta }{/tex}

Recall identity sec² θ – tan²θ = 1

Here, LHS {tex}= \frac { 1 + \sec \theta - \tan \theta } { 1 + \sec \theta + \tan \theta }{/tex}

{tex}= \frac { \sec ^ { 2 } \theta - \tan ^ { 2 } \theta + \sec \theta - \tan \theta } { 1 + \sec \theta + \tan \theta }{/tex}

{tex}= \frac { ( \sec \theta - \tan \theta ) ( \sec \theta + \tan \theta ) + ( \sec \theta - \tan \theta ) } { 1 + \sec \theta + \tan \theta }{/tex} [ because, a² – b² = (a – b)(a + b)]

{tex}= \frac { ( \sec \theta - \tan \theta ) [ \sec \theta + \tan \theta + 1 ] } { ( \sec \theta + \tan \theta + 1 ) }{/tex}

= sec θ – tan θ

{tex}= \frac { 1 } { \cos \theta } - \frac { \sin \theta } { \cos \theta }{/tex}

{tex}= \frac { 1 - \sin \theta } { \cos \theta }{/tex} = RHS

Hence, proved.