To solve this problem, we need to express the given trigonometric expression $\tan^2(x) + 2 \csc(x) \tan(x)$ in terms of $a = \sec(x)$ and $b = \csc(x)$.

Step 1: Simplify the Expression Using Trigonometric Identities

We know: $\sec(x) = \frac{1}{\cos(x)} \quad \text{and} \quad \csc(x) = \frac{1}{\sin(x)}$ Given: $\sec(x) = a \quad \text{and} \quad \csc(x) = b$

Step 2: Express $\tan(x)$ in Terms of $a$ and $b$

$\tan(x) = \frac{\sin(x)}{\cos(x)} = \frac{1}{\csc(x)\sec(x)} = \frac{1}{ab}$ So, $\tan(x) = \frac{1}{ab}$

Step 3: Substitute into the Given Expression

Now substitute $\tan(x)$ and $\csc(x)$ into the expression $\tan^2(x) + 2 \csc(x) \tan(x)$: $\tan^2(x) + 2 \csc(x) \tan(x) = \left(\frac{1}{ab}\right)^2 + 2b\left(\frac{1}{ab}\right)$

Step 4: Simplify the Expression

$\left(\frac{1}{ab}\right)^2 = \frac{1}{a^2b^2}$ $2b\left(\frac{1}{ab}\right) = \frac{2}{a}$ Thus: $\frac{1}{a^2b^2} + \frac{2}{a}$ Finally, find a common denominator: $\frac{1}{a^2b^2} + \frac{2}{a} = \frac{1 + 2a^2b^2}{a^2b^2}$

So, the expression $\tan^2(x) + 2 \csc(x) \tan(x)$ in terms of $a$ and $b$ is: $\boxed{\frac{1 + 2a^2b^2}{a^2b^2}}$

Would you like more details on any of these steps or have any questions?

Here are some related questions:

1. How can we express $\sin(x)$ and $\cos(x)$ in terms of $a$ and $b$?
2. What is the significance of using trigonometric identities in simplifying expressions?
3. How would you approach expressing $\cot(x)$ in terms of $a$ and $b$?
4. Can we generalize this approach to other trigonometric identities?
5. How can these simplifications help in solving trigonometric equations?

Tip: Always remember to double-check the relationships between trigonometric functions when substituting values to ensure accuracy in your expressions.