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Let's solve the given problem:

We are asked to **prove that**:

\[
\frac{\cos x}{1 - \sin x} + \frac{1 - \sin x}{\cos x} \equiv 2 \sec x.
\]

### Proof:

**Step 1: Simplify the left-hand side expression.**

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

We take a common denominator:

\[
\frac{\cos^2 x + (1 - \sin x)^2}{(1 - \sin x) \cos x}.
\]

**Step 2: Expand the numerator.**

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

\[
\cos^2 x + 1 - 2 \sin x + \sin^2 x.
\]

Using the identity \(\cos^2 x + \sin^2 x = 1\), this simplifies to:

\[
1 + 1 - 2 \sin x = 2 - 2 \sin x.
\]

Thus, the expression becomes:

\[
\frac{2 - 2 \sin x}{(1 - \sin x) \cos x}.
\]

**Step 3: Simplify the fraction.**

Factor out 2 from the numerator:

\[
\frac{2(1 - \sin x)}{(1 - \sin x) \cos x}.
\]

Cancel the common factor \(1 - \sin x\) (assuming it is non-zero):

\[
\frac{2}{\cos x}.
\]

**Step 4: Rewrite the result.**

\[
\frac{2}{\cos x} = 2 \sec x.
\]

**Conclusion:**

We have shown that:

\[
\frac{\cos x}{1 - \sin x} + \frac{1 - \sin x}{\cos x} \equiv 2 \sec x.
\]

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Let me know if any step is unclear, or if you need further assistance!

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### 5 Related Questions:
1. What is the identity for \(\sec x\) in terms of \(\cos x\)?
2. How do you factor trigonometric expressions like \(1 - \sin^2 x\)?
3. Can you derive the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\)?
4. How do you find the domain of \(\sec x\)?
5. How would the identity change if the expression was \(\frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x}\)?

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**Tip:** Always try to look for common identities (like Pythagorean identities) when working with trigonometric proofs to simplify expressions easily.

Show that cos(x)/(1 - sin(x)) + (1 - sin(x))/cos(x) ≡ 2 sec(x).

Learn how to prove the trigonometric identity cos(x)/(1 - sin(x)) + (1 - sin(x))/cos(x) ≡ 2 sec(x) using simplification and the Pythagorean identity. This problem is ideal for high school students studying trigonometry.

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