2015/093) Show that $\dfrac{\sec\,A +\tan\, A - 1}{\tan\, A- \sec\, A +1} = \dfrac{\cos\, A}{1-\sin\,A}$

we have
$\dfrac{\sec\,A +\tan\, A - 1}{\tan\, A- \sec\, A +1}$
= $\dfrac{\sec\,A +\tan\, A - (\sec^2A-\tan ^2 A}{\tan\, A- \sec\, A +1}$
=  $\dfrac{\sec\,A +\tan\, A - (\sec\,A+\tan\, A)(\sec\,A-\tan\, A)}{\tan\, A- \sec\, A +1}$
= $\dfrac{(\sec\,A +\tan\, A)(1 - \sec\,A+\tan\, A)}{\tan\, A- \sec\, A +1}$
= $(\sec\,A +\tan\, A)$
= $\dfrac{1}{\cos\, A }+  \dfrac{\sin\, A}{\cos\, A }$
=  $\dfrac{1+ \sin\, A}{\cos\, A }$
=  $\dfrac{(1+ \sin\, A)(1-\sin\, A)}{\cos\, A(1-\sin\, A) }$
= $\dfrac{1-\sin^2A}{\cos\, A(1-\sin\, A) }$
=  $\dfrac{\cos^2A}{\cos\, A(1-\sin\, A) }$
= $\dfrac{\cos\,A}{1-\sin\, A}$