How to prove the trigonometric identity $\frac{1-\sin\varphi}{1+\sin\varphi} = (\sec\varphi - \tan\varphi)^2$ $$\frac{1-\sin\varphi}{1+\sin\varphi}$$
I have no idea how to start this, please help. This problem is essentially supposed to help me solve the proof of  
$$\frac{1-\sin\varphi}{1+\sin\varphi} = (\sec\varphi - \tan\varphi)^2$$
 A: $$(\sec\varphi - \tan\varphi)^2 = \left(\frac{1}{\cos \varphi}-\frac{\sin \varphi}{\cos \varphi}\right)^2 = \frac{(1 - \sin \varphi)^2}{\cos^2 \varphi}$$
$$ = \frac{(1 - \sin \varphi)^2}{ 1 - \sin^2 \varphi} = \frac{(1 - \sin \varphi)^2}{(1 - \sin \varphi)(1+\sin \varphi)} = \frac{1-\sin \varphi}{1+\sin\varphi}$$
A: If we solve the Pythagorean identity 
$$\sin^2\varphi + \cos^2\varphi = 1$$ for $\cos^2\varphi$ 
we obtain 
$$\cos^2\varphi = 1 - \sin^2\varphi$$
Since 
$$\sec\varphi = \frac{1}{\cos\varphi}$$
and 
$$\tan\varphi = \frac{\sin\varphi}{\cos\varphi}$$
we need to obtain $\cos^2\varphi$ in the denominator.  We can do this by multiplying the denominator by $1 - \sin\varphi$ to obtain $1 - \sin^2\varphi = \cos^2\varphi$.  However, what we do to the denominator, we must also do to the numerator in order to preserve equality. 
\begin{align*}
\frac{1 - \sin\varphi}{1 + \sin\varphi} & = \frac{1 - \sin\varphi}{1 + \sin\varphi} \cdot \frac{1 - \sin\varphi}{1 - \sin\varphi}\\
& = \frac{1 - 2\sin\varphi + \sin^2\varphi}{1 - \sin^2\varphi}\\
& = \frac{1 - 2\sin\varphi + \sin^2\varphi}{\cos^2\varphi}\\
& = \frac{1}{\cos^2\varphi} - \frac{2\sin\varphi}{\cos^2\varphi} + \frac{\sin^2\varphi}{\cos^2\varphi}\\
& = \sec^2\varphi - 2\frac{\sin\varphi}{\cos\varphi} \cdot \frac{1}{\cos\varphi} + \tan^2\varphi\\
& = \sec^2\varphi - 2\tan\varphi\sec\varphi + \tan^2\varphi\\
& = (\sec\varphi - \tan\varphi)^2
\end{align*}
A: Multiply by 
$$\frac{1+\sin\theta}{1+\sin\theta}$$
