The given expression is:

$\frac{5\tan^2(x)}{\sec(x) - 1}$

We can simplify this step by step.

Step 1: Rewrite in terms of sine and cosine

Recall the trigonometric identities: $\tan(x) = \frac{\sin(x)}{\cos(x)} \quad \text{and} \quad \sec(x) = \frac{1}{\cos(x)}.$

Substituting these identities: $\frac{5\tan^2(x)}{\sec(x) - 1} = \frac{5\left(\frac{\sin^2(x)}{\cos^2(x)}\right)}{\frac{1}{\cos(x)} - 1}.$

Step 2: Simplify the denominator

$\frac{1}{\cos(x)} - 1 = \frac{1 - \cos(x)}{\cos(x)}.$

So the expression becomes: $\frac{5\left(\frac{\sin^2(x)}{\cos^2(x)}\right)}{\frac{1 - \cos(x)}{\cos(x)}}.$

Step 3: Simplify the fraction

Dividing by a fraction is equivalent to multiplying by its reciprocal: $\frac{5\left(\frac{\sin^2(x)}{\cos^2(x)}\right)}{\frac{1 - \cos(x)}{\cos(x)}} = 5 \cdot \frac{\sin^2(x)}{\cos^2(x)} \cdot \frac{\cos(x)}{1 - \cos(x)}.$

Cancel one $\cos(x)$ in the numerator and denominator: $= \frac{5\sin^2(x)}{\cos(x)(1 - \cos(x))}.$

Final Expression:

The simplified form of the given expression is: $\frac{5\sin^2(x)}{\cos(x)(1 - \cos(x))}.$

Let me know if you would like further explanation or additional steps.

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Questions for deeper understanding:

1. What are the key trigonometric identities used in this solution?
2. How does dividing by a fraction simplify into multiplication by the reciprocal?
3. Why do we replace $\sec(x)$ and $\tan(x)$ with their definitions in terms of sine and cosine?
4. How can this expression be evaluated for specific values of $x$?
5. What conditions must be true for this expression to be defined?

Tip:

Always check for domain restrictions in trigonometric expressions—like division by zero or undefined values of functions like $\sec(x)$ or $\tan(x)$.