Trigonometry

Using the identities

$\sin(x) = 2\sin(\frac{x}{2})\cos(\frac{x}{2})$ ,

$\cos(x) = 1 - 2\sin^{2}(\frac{x}{2}) \Longrightarrow 1 - \cos(x) = 2\sin^{2}(\frac{x}{2})$ and

$\cos(x) = 2\cos^{2}(\frac{x}{2}) - 1 \Longrightarrow 1 + \cos(x) = 2\cos^{2}(\frac{x}{2})$ , we see that

$\dfrac{1 + \sin(x) - \cos(x)}{1 + \sin(x) + \cos(x)} = \dfrac{(1 - \cos(x)) + \sin(x)}{(1 + \cos(x)) + \sin(x)} = \dfrac{2\sin^{2}(\frac{x}{2}) + 2\sin(\frac{x}{2})\cos(\frac{x}{2})}{2\cos^{2}(\frac{x}{2}) + 2\sin(\frac{x}{2})\cos(\frac{x}{2})} =$

$\dfrac{2\sin(\frac{x}{2})(\sin(\frac{x}{2}) + \cos(\frac{x}{2}))}{2\cos(\frac{x}{2})(\cos(\frac{x}{2}) + \sin(\frac{x}{2}))} = \dfrac{\sin(\frac{x}{2})}{\cos(\frac{x}{2})} = \boxed{\tan(\frac{x}{2})}$ .

Using half-angle tangent substitution or Weierstrass substitution and let $t = \tan \dfrac x2$ , we have:

$\begin{aligned} \frac {1+\sin x - \cos x}{1+\sin x + \cos x} & = \frac {1+\frac {2t}{1+t^2}-\frac {1-t^2}{1+t^2}}{1+\frac {2t}{1+t^2}+\frac {1-t^2}{1+t^2}} \\ & = \frac {1+t^2 + 2t - 1+t^2}{1+t^2 + 2t + 1-t^2} \\ & = \frac {2t +2t^2}{2+2t} \\ & = t = \boxed{\tan \dfrac x2} \end{aligned}$