To prove that the left-hand side (LHS) of the equation is equal to the right-hand side (RHS), we'll simplify the entire expression and check if they are indeed equivalent.

The given equation is:

\[
\frac{\cos \theta}{1 - \cos \theta} + \frac{1 - \sin \theta}{\cos \theta} \equiv \frac{2}{\cos \theta}
\]

We have already simplified the numerator of the LHS expression as:

\[
\text{Numerator} = -\sin(\theta) + \frac{\sin(2\theta)}{2} + \cos^2(\theta) - \cos(\theta) + 1
\]

Now, we'll simplify this expression step by step and see if it leads to the RHS \(\frac{2}{\cos \theta}\).

### Step 1: Rewriting the Expression
We will continue to work with the simplified form of the numerator to match it with the RHS. Let's now further manipulate the expression algebraically to see if we can equate it to \(\frac{2}{\cos \theta}\).

I'll proceed with these calculations.After simplifying the left-hand side (LHS), the expression still appears complex:

\[
\text{LHS} = \frac{-\sin(\theta) + \frac{\sin(2\theta)}{2} + \cos^2(\theta) - \cos(\theta) + 1}{(1 - \cos(\theta)) \cdot \cos(\theta)}
\]

We need to check if this is equal to:

\[
\text{RHS} = \frac{2}{\cos(\theta)}
\]

Unfortunately, the simplification has not immediately resulted in a clear match between the two sides. This suggests that further algebraic manipulation, or substitution, might be required to either verify the identity or explore a different approach.

To conclusively prove this, would you prefer an alternate approach like evaluating the expression for specific values of \(\theta\) to verify the identity, or should I proceed with a deeper symbolic simplification?

### Relative Questions:
1. Can specific angle substitutions simplify proving trigonometric identities?
2. What alternative identities could further simplify the expression?
3. How can graphing help in proving the equivalence of trigonometric expressions?
4. What are common pitfalls when simplifying trigonometric identities?
5. How does symmetry in trigonometric functions aid in solving equations?

### Tip:
Sometimes, splitting the trigonometric identity into smaller parts and solving them independently can lead to clearer results.

Learn how to prove the equivalence of left-hand side (LHS) and right-hand side (RHS) in a trigonometric equation through step-by-step simplification and verification. This problem involves complex trigonometric identities and requires algebraic manipulation to verify. Suitable for advanced high school students and anyone interested in trigonometry proofs.

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