We are asked to prove the trigonometric identity:

$\frac{\csc x}{\cot x + \tan x} = \cos x$

Step-by-step proof:

1. Express in terms of basic trigonometric functions:

   Recall the definitions of cosecant, cotangent, and tangent in terms of sine and cosine: $\csc x = \frac{1}{\sin x}, \quad \cot x = \frac{\cos x}{\sin x}, \quad \tan x = \frac{\sin x}{\cos x}$

   Substitute these into the left-hand side (LHS): $\frac{\csc x}{\cot x + \tan x} = \frac{\frac{1}{\sin x}}{\frac{\cos x}{\sin x} + \frac{\sin x}{\cos x}}$

2. Simplify the denominator:

   To simplify $\frac{\cos x}{\sin x} + \frac{\sin x}{\cos x}$, get a common denominator: $\frac{\cos x}{\sin x} + \frac{\sin x}{\cos x} = \frac{\cos^2 x + \sin^2 x}{\sin x \cos x}$

   Use the Pythagorean identity $\cos^2 x + \sin^2 x = 1$: $\frac{1}{\sin x \cos x}$

3. Rewrite the expression:

   Now, substitute this simplified denominator back into the expression: $\frac{\frac{1}{\sin x}}{\frac{1}{\sin x \cos x}} = \frac{1}{\sin x} \cdot \frac{\sin x \cos x}{1} = \cos x$

Thus, the left-hand side simplifies to $\cos x$, which is exactly the right-hand side (RHS).

Conclusion:

$\boxed{\frac{\csc x}{\cot x + \tan x} = \cos x}$

This proves the given identity.

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Would you like more details on any step, or have any other questions?

Here are 5 related questions to expand on this:

1. How do we derive the Pythagorean identity $\cos^2 x + \sin^2 x = 1$?
2. What is the relationship between sine, cosine, and the unit circle?
3. How can we prove $\csc x = \frac{1}{\sin x}$ using the unit circle?
4. How can we solve trigonometric equations involving both sine and cosine functions?
5. What are the conditions under which trigonometric identities hold?

Tip: When dealing with trigonometric identities, expressing all functions in terms of sine and cosine often simplifies the problem.