problem 2014/109) prove that $\dfrac{\sin\,x - \cos\, x + 1}{\sin\, x + \cos\, x – 1} = \dfrac{1 + \tan \frac{x}{2}}{1- \tan\frac{x}{2}}$

Proof

convert it into $\dfrac{x}{2}$ form

numerator = $2 sin \dfrac{x}{2}\cos\dfrac{x}{2} + 2 \sin ^2\dfrac{x}{2}$ (as $\cos\, x= 1 - 2 \sin ^2 \dfrac{x}{2}$) = $2 \sin \dfrac{x}{2}( \cos \dfrac{x}{2} + \sin \dfrac{x}{2})$
denominator = $2 \sin \dfrac{x}{2} \cos \dfrac{x}{2} - 2 \cos^2 \dfrac{x}{2}$ (as $cos\, x= 2 \cos^2 \dfrac{x}{2} - 1$ = $2 \sin \dfrac{x}{2}(\cos \dfrac{x}{2}- \sin \dfrac{x}{2})$
dividing
we get
$\dfrac{\sin\,x - \cos\,x + 1}{\sin\,x + \cos\,x - 1} = \dfrac{\cos \dfrac{x}{2} + \sin \dfrac{x}{2}}{\cos\dfrac{x}{2}-\sin \dfrac{x}{2}}$
= $\dfrac{(1 + \tan \dfrac{x}{2})\cos\dfrac{x}{2}}{(1 - \tan \dfrac{x}{2})\cos\dfrac{x}{2}}$
= $\dfrac{1 + \tan \dfrac{x}{2}}{(1 - \tan \dfrac{x}{2}}$