sin-cos+1÷sin+cos-1=1÷sec-tan

We have to prove that,{tex}\frac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} = \frac{1}{{\sec \theta - \tan \theta }}{/tex} using identity {tex}sec^2\theta=1+tan^2\theta{/tex}

LHS = {tex}\frac{{\sin \theta - \cos \theta + 1}}{{\sin \theta + \cos \theta - 1}} {/tex}{tex} = \frac{{\tan \theta - 1 + \sec \theta }}{{\tan \theta + 1 - \sec \theta }}{/tex} [  dividing the numerator and denominator by {tex}\cos{\theta}{/tex}.]

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

{tex}=\frac{{(\tan^2\theta-\sec^2\theta)-}(tan\theta-\sec\theta)}{\{{(\tan\theta-\sec\theta)+1\}}(\tan\theta-\sec\theta)}{/tex}    [{tex}\because (a-b)(a+b)=a^2-b^2{/tex}]

{tex} = \frac{{-1-\tan \theta + \sec \theta }}{{(\tan \theta - \sec \theta+1)(\tan{\theta}-\sec{\theta}) }}{/tex}[{tex}\because \tan^2\theta-\sec^2\theta=-1{/tex}]

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

{tex} = \frac{1}{{\sec \theta - \tan \theta }}{/tex}=RHS

Hence Proved.