Using Lagrange's mean value theorem prove that if `b gt a gt 0`
`"then " (b-a)/(1+b^(2)) lt tan^(-1) b -tan^(-1) a lt (b-a)/(1+a^(2))`

Explanation

To prove the inequality \(\frac{b-a}{1+b^2} < \tan^{-1} b - \tan^{-1} a < \frac{b-a}{1+a^2}\) using Lagrange's Mean Value Theorem, we will follow these steps:

### Step 1: Define the function
Let \( f(x) = \tan^{-1} x \). We need to analyze this function on the interval \([a, b]\) where \( b > a > 0 \).

### Step 2: Check the conditions for Mean Value Theorem
The function \( f(x) \) is continuous and differentiable for all \( x \) in \((0, \infty)\). Thus, the conditions for applying Lagrange's Mean Value Theorem are satisfied.

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