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

Question

anonymous · Answer

$$\frac{ \sec \theta+\tan \theta-1 }{ \ \tan  \theta -\ \sec   \theta  +1 }=\frac{ \cos \theta  }{ 1-\sin \theta  }$$
L.H.S=$$\frac{ \sec \theta  +\tan \theta-1 }{ \tan \theta -\sec \theta+1 }=\frac{ \sec \theta +\tan \theta -\left( \sec ^2\theta-\tan ^2\theta \right) }{ \tan \theta-\sec \theta  +1 }$$
$$=\frac{ \sec \theta+\tan \theta-\left( \sec \theta +\tan \theta \right)\left( \sec \theta -\tan \theta   \right) }{ \tan  \theta-\sec  \theta +1}$$
$$=\frac{ \left( \sec \theta +\tan \theta  \right)\left( 1-\sec \theta+\tan  \theta \right) }{ \tan \theta -\sec \theta+1 }$$
$$=(\sec \theta+\tan \theta)*\frac{ \sec \theta  -\tan \theta  }{ \sec \theta -\tan \theta }=\frac{ \sec ^2\theta -\tan ^2 \theta}{   \sec \theta -\tan \theta }$$
$$=\frac{ 1 }{ \sec \theta  -\tan \theta }=?$$