https://file.aiquickdraw.com/mathbot/file/1726112971628asa0gf2p.jpg

Let's break down the problem into two parts as described in the image:

### Part 1: Prove the following inequality for \(a < b\):
\[
\frac{b-a}{1+b^2} < (\tan^{-1} b - \tan^{-1} a) < \frac{b-a}{1+a^2}.
\]

This inequality is an application of the **Mean Value Theorem** (MVT) for the function \( f(x) = \tan^{-1}(x) \).

#### Applying the Mean Value Theorem:
1. The Mean Value Theorem states that for a continuous and differentiable function \( f(x) \) on the interval \([a, b]\), there exists some \( c \in (a, b) \) such that:
   \[
   f'(c) = \frac{f(b) - f(a)}{b - a}.
   \]
2. The derivative of \( f(x) = \tan^{-1}(x) \) is:
   \[
   f'(x) = \frac{1}{1 + x^2}.
   \]
3. Therefore, applying the MVT to \( f(x) = \tan^{-1}(x) \) on the interval \([a, b]\), we get:
   \[
   \frac{\tan^{-1}(b) - \tan^{-1}(a)}{b - a} = \frac{1}{1 + c^2},
   \]
   where \( c \in (a, b) \).

#### Inequality Bound:
1. Since \( c \in (a, b) \), the value of \( \frac{1}{1 + c^2} \) lies between \( \frac{1}{1 + b^2} \) and \( \frac{1}{1 + a^2} \).
2. This gives us the inequality:
   \[
   \frac{1}{1 + b^2} < \frac{1}{1 + c^2} < \frac{1}{1 + a^2}.
   \]
3. Multiplying through by \( b - a \), we get:
   \[
   \frac{b - a}{1 + b^2} < \tan^{-1}(b) - \tan^{-1}(a) < \frac{b - a}{1 + a^2}.
   \]
This completes the proof of the inequality.

---

### Part 2: Deduce the inequality
\[
\frac{5\pi + 4}{20} < \tan^{-1}(2) < \frac{\pi + 2}{4}.
\]

To deduce this result, we can use the inequality from the first part and substitute appropriate values for \(a\) and \(b\).

#### Substitution:
1. Let \(a = 1\) and \(b = 2\). Then, applying the inequality from Part 1, we get:
   \[
   \frac{2 - 1}{1 + 2^2} < \tan^{-1}(2) - \tan^{-1}(1) < \frac{2 - 1}{1 + 1^2}.
   \]
2. Simplifying both sides:
   \[
   \frac{1}{5} < \tan^{-1}(2) - \frac{\pi}{4} < \frac{1}{2}.
   \]
3. Adding \( \frac{\pi}{4} \) to all parts of the inequality:
   \[
   \frac{1}{5} + \frac{\pi}{4} < \tan^{-1}(2) < \frac{1}{2} + \frac{\pi}{4}.
   \]
4. Simplifying both expressions:
   \[
   \frac{5\pi + 4}{20} < \tan^{-1}(2) < \frac{\pi + 2}{4}.
   \]
This is the required inequality.

---

Would you like to explore any specific part in more detail? Let me know if you have questions, or if you'd like me to explain any step more thoroughly!

Here are 5 relative questions that expand on the above information:
1. How does the Mean Value Theorem apply to functions beyond \(\tan^{-1}(x)\)?
2. Can this method be extended to other inverse trigonometric functions like \(\sin^{-1}(x)\)?
3. What is the significance of the function \(\tan^{-1}(x)\) in calculus and analysis?
4. Why do we use specific values like \(a = 1\) and \(b = 2\) for deducing inequalities?
5. How does understanding inequalities help in estimating values of inverse trigonometric functions?

**Tip:** The Mean Value Theorem is a powerful tool for proving inequalities involving derivatives and differences between function values—try applying it to different functions!

If a < b, prove that (b-a)/(1+b^2) < (tan^(-1)b - tan^(-1)a) < (b-a)/(1+a^2). Hence deduce that (5π + 4)/20 < tan^(-1)(2) < (π + 2)/4.

In this problem, we prove an inequality involving arctangent using the Mean Value Theorem and deduce bounds for tan^(-1)(2). This exercise covers important calculus concepts, including inverse trigonometric functions and inequalities.

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