The given expression is:

$\frac{\cot x + \csc x}{\sin x + \tan x}$

You are asked to find which of the following expressions matches this identity from the provided options:

A. $\sec x \cot x$
B. $\csc x \cot x$
C. $\cos x \sin x$
D. $\tan^2 x - 1$

Step-by-step Solution:

Step 1: Simplify the numerator and denominator separately

• The numerator: $\cot x + \csc x$

  • Recall that $\cot x = \frac{\cos x}{\sin x}$ and $\csc x = \frac{1}{\sin x}$, so: $\cot x + \csc x = \frac{\cos x}{\sin x} + \frac{1}{\sin x} = \frac{\cos x + 1}{\sin x}$

• The denominator: $\sin x + \tan x$

  • Recall that $\tan x = \frac{\sin x}{\cos x}$, so: $\sin x + \tan x = \sin x + \frac{\sin x}{\cos x} = \frac{\sin x \cos x + \sin x}{\cos x} = \frac{\sin x (\cos x + 1)}{\cos x}$

Step 2: Substitute the simplified forms into the original expression

$\frac{\cot x + \csc x}{\sin x + \tan x} = \frac{\frac{\cos x + 1}{\sin x}}{\frac{\sin x (\cos x + 1)}{\cos x}} = \frac{\cos x + 1}{\sin x} \times \frac{\cos x}{\sin x (\cos x + 1)}$

Step 3: Cancel common terms

The $\cos x + 1$ cancels out from both the numerator and denominator, leaving:

$\frac{\cos x}{\sin^2 x}$

Step 4: Recognize the final form

We know that $\frac{\cos x}{\sin^2 x} = \cot x \csc x$.

Thus, the correct answer is Option B: $\csc x \cot x$.

Would you like more details or have any questions? Here are some related questions to explore further:

1. What is the relationship between the trigonometric functions $\cot x$ and $\csc x$?
2. How do you derive the identity $\tan^2 x - 1 = \sec^2 x$?
3. What are the steps for simplifying complex trigonometric expressions?
4. How does the Pythagorean identity relate to other trigonometric identities?
5. What is the significance of the angle sum and difference identities in simplifying expressions?

Tip: Always check for opportunities to cancel common terms when simplifying trigonometric expressions.