You provided the following expression:

$\tan^{-1} \left( \frac{\cos x}{1 - \sin x} \right)$

This expression can be simplified using a trigonometric identity. Let's work through the simplification.

Step-by-step Simplification:

1. Recognize the tangent identity: $\tan\left(\frac{\theta}{2}\right) = \frac{1 - \cos\theta}{\sin\theta}$

2. Substitute the complementary angle identity for the tangent: Using $\tan^{-1}(\cdot)$, we can rewrite the expression inside the inverse tangent as: $\frac{\cos x}{1 - \sin x}$ This is equivalent to $\tan\left(\frac{\pi}{2} - x\right)$, based on the known trigonometric identity: $\frac{\cos x}{1 - \sin x} = \tan\left(\frac{x}{2}\right)$

3. Final result: Hence, the original expression simplifies to: $\tan^{-1}\left( \frac{\cos x}{1 - \sin x} \right) = \frac{x}{2}$

So, the simplified form of the given expression is:

$\boxed{\frac{x}{2}}$

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Would you like more details on any step? Let me know if you have questions!

Here are 5 related questions for further exploration:

1. How is the arctangent function related to inverse trigonometric functions in general?
2. What are other trigonometric identities that involve arctan?
3. Can you solve an equation involving $\tan^{-1}$?
4. How does the simplification of trigonometric expressions benefit calculus?
5. What is the geometric interpretation of $\tan^{-1}$?

Tip: Understanding the relationships between the different trigonometric functions can simplify complex expressions significantly!