<s>[INST] Need help in proving that $\frac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}$ We need to prove that $$\dfrac{\sin\theta - \cos\theta + 1}{\sin\theta + \cos\theta - 1} = \frac 1{\sec\theta - \tan\theta}$$
I have tried and it gets confusing.
 [/INST] These kind of questions often benefit from the identity $a^2-b^2=(a-b)(a+b)$ in conjunction with Pythagorean trig identitities. Here,
$$
\begin{align}
\frac{\sin t-\cos t+1}{\sin t+\cos t -1}&=\frac{(\sin t+1)-\cos t}{(\sin t+\cos t) -1}\cdot\frac{(\sin t+1)+\cos t}{(\sin t+\cos t) +1}\\
&=\frac{\sin^2 t+2\sin t+1-\cos^2 t}{\sin^2 t+2\sin t\cos t+\cos^2 t -1}\\
&=\frac{2\sin^2 t+2\sin t}{2\sin t\cos t}\\
&=\frac{\sin t+1}{\cos t}\\
&=\tan t+\sec t\\
&=(\tan t+\sec t)\frac{\tan t-\sec t}{\tan t-\sec t}\\
&=\frac{\tan^2 t-\sec^2 t}{\tan t-\sec t}\\
&=\frac{1}{\sec t-\tan t}\\
\end{align}
$$
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