We need to prove the identity:

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

### Step 1: Express everything in terms of \(\sin x\) and \(\cos x\).
- We know that \(\sec x = \frac{1}{\cos x}\) and \(\tan x = \frac{\sin x}{\cos x}\).
- The right-hand side (RHS) can be rewritten as:

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

Thus, the equation to prove becomes:

\[
\frac{1 - \sin x}{\cos x} = \frac{\cos x}{1 + \sin x}
\]

### Step 2: Cross-multiply to eliminate the denominators.
Multiplying both sides of the equation by \((\cos x)(1 + \sin x)\), we get:

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

### Step 3: Simplify both sides.
- On the left-hand side, we use the difference of squares formula:

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

- On the right-hand side, \((\cos x)^2\) remains \(\cos^2 x\).

Now the equation is:

\[
1 - \sin^2 x = \cos^2 x
\]

### Step 4: Use the Pythagorean identity.
From the Pythagorean identity, we know:

\[
1 - \sin^2 x = \cos^2 x
\]

Thus, both sides of the equation are equal:

\[
\cos^2 x = \cos^2 x
\]

### Conclusion:
The given identity is proven to be true.

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

---

Do you want further details or have any questions? Here are 5 related questions for practice:

1. Prove the identity \(\tan x = \frac{\sin x}{\cos x}\).
2. Simplify \(\frac{\sin^2 x + \cos^2 x}{\sin^2 x}\).
3. Verify the identity \(\frac{\sin x}{1 - \cos x} = \frac{1 + \cos x}{\sin x}\).
4. Solve \(\sec x + \tan x = 2\) for \(x\).
5. Prove \(\sin 2x = 2 \sin x \cos x\).

**Tip:** When working with trigonometric identities, always start by expressing all functions in terms of \(\sin x\) and \(\cos x\) to simplify comparisons.

Step-by-step solution to proving the trigonometric identity 1 - sinx/cosx = 1/(secx + tanx), using key trigonometric concepts such as secant, tangent, and the Pythagorean identity. Suitable for high school students in Grades 10-12.

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