Please prove $\frac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan \left(\frac \theta 2\right)$ 
Prove that $\dfrac{1 + \sin\theta - \cos\theta}{1 + \sin\theta + \cos\theta} = \tan\left(\dfrac{\theta}{2}\right)$

Also it is a question of S.L. Loney's Plane Trignonometry  
What I've tried by now:  
\begin{align}
& =\frac{1+\sin\theta-\sin(90-\theta)}{1+\cos(90-\theta)+\cos\theta} \\[10pt]
& =\frac{1+2\cos45^\circ \sin(\theta-45^\circ)}{1+2\cos45^\circ \cos(45-\theta)} \end{align}  

Cause I do know
  \begin{align} & \sin c + \sin d = 2\sin\left(\frac{c+d}{2}\right)\cos\left(\frac{c-d}{2}\right) \\[10pt] \text{and } & \cos c + \cos d = 2\cos\left(\frac{c+d}{2}\right)\sin\left(\frac{c-d}{2}\right)
\end{align}

I can't think of what to do next..

Let $\theta = 2 \phi$, then the thing to be proven is:

Prove that $$\frac{1 + \sin(2\phi) - \cos(2\phi)}{1 + \sin(2\phi) + \cos(2\phi)} = \tan(\phi)$$

Then use:
$$\sin(2\phi) = 2 \sin \phi \cos \phi$$
$$\cos(2\phi) = \cos^2 \phi - \sin^2 \phi$$
and:
$$\sin^2 \phi + \cos^2 \phi = 1$$
