Using the trigonometric identity sec^2θ = 1 + tan^2θ, prove that (sin θ - cos θ + 1) / (sin θ + cos θ - 1) = 1 / (sec θ - tan θ).

To prove the given equation:

\[
\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{1}{\sec \theta - \tan \theta}
\]

using the trigonometric identity \(\sec^2 \theta = 1 + \tan^2 \theta\), follow these steps:

### Step 1: Express the right-hand side in terms of sine and cosine.

Start with the right-hand side of the equation:

\[
\frac{1}{\sec \theta - \tan \theta}
\]

We know that:

\[
\sec \theta = \frac{1}{\cos \theta} \quad \text{and} \quad \tan \theta = \frac{\sin \theta}{\cos \theta}
\]

Thus:

\[
\sec \theta - \tan \theta = \frac{1}{\cos \theta} - \frac{\sin \theta}{\cos \theta}
\]

Now, factor out the common denominator \(\cos \theta\):

\[
\sec \theta - \tan \theta = \frac{1 - \sin \theta}{\cos \theta}
\]

Therefore, the right-hand side becomes:

\[
\frac{1}{\sec \theta - \tan \theta} = \frac{\cos \theta}{1 - \sin \theta}
\]

### Step 2: Simplify the left-hand side.

Now, let's simplify the left-hand side:

\[
\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1}
\]

We want to simplify this expression in a way that matches the right-hand side. Notice that the structure of both sides looks similar, so we will attempt to manipulate it using trigonometric identities.

### Step 3: Try multiplying numerator and denominator by a conjugate.

We will multiply both the numerator and denominator of the left-hand side by the conjugate of the denominator \(\sin \theta + \cos \theta - 1\), which is \(\sin \theta + \cos \theta + 1\):

\[
\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} \times \frac{\sin \theta + \cos \theta + 1}{\sin \theta + \cos \theta + 1}
\]

This gives:

\[
\frac{(\sin \theta - \cos \theta + 1)(\sin \theta + \cos \theta + 1)}{(\sin \theta + \cos \theta - 1)(\sin \theta + \cos \theta + 1)}
\]

### Step 4: Simplify the denominator.

The denominator is a difference of squares:

\[
(\sin \theta + \cos \theta - 1)(\sin \theta + \cos \theta + 1) = (\sin \theta + \cos \theta)^2 - 1^2
\]

Now, expand \((\sin \theta + \cos \theta)^2\):

\[
(\sin \theta + \cos \theta)^2 = \sin^2 \theta + 2 \sin \theta \cos \theta + \cos^2 \theta
\]

Since \(\sin^2 \theta + \cos^2 \theta = 1\), this becomes:

\[
1 + 2 \sin \theta \cos \theta
\]

So the denominator simplifies to:

\[
(1 + 2 \sin \theta \cos \theta) - 1 = 2 \sin \theta \cos \theta
\]

### Step 5: Simplify the numerator.

Now, expand the numerator:

\[
(\sin \theta - \cos \theta + 1)(\sin \theta + \cos \theta + 1)
\]

Use distributive property to expand:

\[
= \sin \theta (\sin \theta + \cos \theta + 1) - \cos \theta (\sin \theta + \cos \theta + 1) + 1 (\sin \theta + \cos \theta + 1)
\]

Expanding each term:

\[
= \sin^2 \theta + \sin \theta \cos \theta + \sin \theta - \cos \theta \sin \theta - \cos^2 \theta - \cos \theta + \sin \theta + \cos \theta + 1
\]

Simplify the expression:

\[
= \sin^2 \theta - \cos^2 \theta + 2 \sin \theta + 1
\]

Now, use the identity \(\sin^2 \theta - \cos^2 \theta = -\cos 2\theta\), but we can also directly work with the form:

\[
= 1 - \cos^2 \theta + 2 \sin \theta + 1
\]

### Step 6: Equating both sides.

Now, combining everything we get:

\[
\frac{\sin \theta - \cos \theta + 1}{\sin \theta + \cos \theta - 1} = \frac{cos