If ` A = 1/1 cot ^(-1) (1/1) + 1/2 cot^(-1) (1/2) + 1/3 cot ^(-1) ( 1/3) " and " B = 1 cot^(-1) ( 1) + 2 cot^(-1) (2) + 3 cot^(-1) (3) " then " |B - A| " is equal to " (api)/b + c/d cot ^(-1) (3) `
where `a,b,c,d in N` are in their lowest form , find ` ( b -a - c - d) `

Answer

To solve the problem, we need to evaluate the expressions for \( A \) and \( B \) and then find \( |B - A| \).

### Step 1: Calculate \( A \)

Given:
\[ A = \frac{1}{1} \cot^{-1}(1) + \frac{1}{2} \cot^{-1}\left(\frac{1}{2}\right) + \frac{1}{3} \cot^{-1}\left(\frac{1}{3}\right) \]

Calculating each term:
- \( \cot^{-1}(1) = \frac{\pi}{4} \)
- \( \cot^{-1}\left(\frac{1}{2}\right) = \tan^{-1}(2) \) (since \( \cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) \))
- \( \cot^{-1}\left(\frac{1}{3}\right) = \tan^{-1}(3) \)

Thus,
\[ A = 1 \cdot \frac{\pi}{4} + \frac{1}{2} \tan^{-1}(2) + \frac{1}{3} \tan^{-1}(3) \]

### Step 2: Simplify \( A \)

Now, substituting the values:
\[ A = \frac{\pi}{4} + \frac{1}{2} \tan^{-1}(2) + \frac{1}{3} \tan^{-1}(3) \]

### Step 3: Calculate \( B \)

Given:
\[ B = 1 \cdot \cot^{-1}(1) + 2 \cdot \cot^{-1}(2) + 3 \cdot \cot^{-1}(3) \]

Calculating each term:
- \( \cot^{-1}(1) = \frac{\pi}{4} \)
- \( \cot^{-1}(2) = \tan^{-1}\left(\frac{1}{2}\right) \)
- \( \cot^{-1}(3) = \tan^{-1}\left(\frac{1}{3}\right) \)

Thus,
\[ B = \frac{\pi}{4} + 2 \tan^{-1}\left(\frac{1}{2}\right) + 3 \tan^{-1}\left(\frac{1}{3}\right) \]

### Step 4: Simplify \( B \)

Now substituting the values:
\[ B = \frac{\pi}{4} + 2 \tan^{-1}\left(\frac{1}{2}\right) + 3 \tan^{-1}\left(\frac{1}{3}\right) \]

### Step 5: Find \( |B - A| \)

Now we compute \( |B - A| \):
\[ |B - A| = \left| \left( \frac{\pi}{4} + 2 \tan^{-1}\left(\frac{1}{2}\right) + 3 \tan^{-1}\left(\frac{1}{3}\right) \right) - \left( \frac{\pi}{4} + \frac{1}{2} \tan^{-1}(2) + \frac{1}{3} \tan^{-1}(3) \right) \right| \]

This simplifies to:
\[ |B - A| = \left| 2 \tan^{-1}\left(\frac{1}{2}\right) + 3 \tan^{-1}\left(\frac{1}{3}\right) - \frac{1}{2} \tan^{-1}(2) - \frac{1}{3} \tan^{-1}(3) \right| \]

### Step 6: Use properties of inverse tangent

Using the property \( \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x + y}{1 - xy}\right) \) for appropriate values, we can combine terms.

After simplification, we find:
\[ |B - A| = \frac{5\pi}{24} + \frac{5}{6} \cot^{-1}(3) \]

### Step 7: Compare with the given form

The problem states:
\[ |B - A| = \frac{a\pi}{b} + \frac{c}{d} \cot^{-1}(3) \]

From our result:
- \( a = 5 \)
- \( b = 24 \)
- \( c = 5 \)
- \( d = 6 \)

### Step 8: Calculate \( b - a - c - d \)

Now we calculate:
\[ b - a - c - d = 24 - 5 - 5 - 6 = 8 \]

### Final Answer

Thus, the final answer is:
\[ \boxed{8} \]