Trigonometric equality: $\frac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \frac{A}{2}$ Can you guys give me a hint on how to proceed with proving this trigonometric equality? I have a feeling I need to use the half angle identity for $\tan \frac{\theta}{2}$. The stuff I have tried so far(multiplying numerator and denominator by $1 + \sin A - \cos A$) has lead to a dead end.

Prove that,
  $$
\dfrac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \dfrac{A}{2}
$$


There is a theorem$^1$ that is worth knowing (I used it in this answer to this question) that states: 
All direct trigonometric functions of $A$ ($2\alpha$) can be expressed rationally in terms of the tangent of $\frac{A}{2}$ ($\alpha$). 
Combining $\sin A=2\sin \frac{A}{2}\cdot \cos \frac{A}{2}$ and $\cos ^{2}
\frac{A}{2}+\sin ^{2}\frac{A}{2}=1$, we get (for $\cos \frac{A}{2}\neq 0$)
$$\sin A=\frac{2\sin \frac{A}{2}\cdot \cos \frac{A}{2}}{\cos ^{2}\frac{A}{2}
+\sin ^{2}\frac{A}{2}}=\frac{2\tan \frac{A}{2}}{1+\tan ^{2}\frac{A}{2}};
\qquad (1)$$
and from $\cos A=\cos ^{2}\frac{A}{2}-\sin ^{2}\frac{A}{2}$, for $\cos \frac{A}{2}\neq 0$,
$$\cos A=\frac{\cos ^{2}\frac{A}{2}-\sin ^{2}\frac{A}{2}}{\cos ^{2}\frac{A}{2}
+\sin ^{2}\frac{A}{2}}=\frac{1-\tan ^{2}\frac{A}{2}}{1+\tan ^{2}\frac{A}{2}}.
\qquad (2)$$
To prove the identity multiply the LHS numerator and denominator by $1+\tan
^{2}\frac{A}{2}$ and use $(1)$ and $(2)$:
$$\begin{eqnarray*}
\frac{1+\sin A-\cos A}{1+\sin A+\cos A} &=&\frac{1+2\tan \frac{A}{2}-1+\tan
^{2}\frac{A}{2}}{1+2\tan \frac{A}{2}+1+\tan ^{2}\frac{A}{2}} \\
&=&\frac{2\tan ^{2}\frac{A}{2}+2\tan \frac{A}{2}}{2+2\tan \frac{A}{2}}=\tan 
\frac{A}{2}.
\end{eqnarray*}$$
--
$^1$ J. Calado, Compêndio de Trigonometria, 1967.
