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:

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

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

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 $\frac{\cos x}{1 + \sin x} + \frac{1 + \sin x}{\cos x}$?

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