<s>[INST] 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  
$$\frac{1-\sin\varphi}{1+\sin\varphi} = (\sec\varphi - \tan\varphi)^2$$
 [/INST] $$(\sec\varphi - \tan\varphi)^2 = \left(\frac{1}{\cos \varphi}-\frac{\sin \varphi}{\cos \varphi}\right)^2 = \frac{(1 - \sin \varphi)^2}{\cos^2 \varphi}$$
$$ = \frac{(1 - \sin \varphi)^2}{ 1 - \sin^2 \varphi} = \frac{(1 - \sin \varphi)^2}{(1 - \sin \varphi)(1+\sin \varphi)} = \frac{1-\sin \varphi}{1+\sin\varphi}$$
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