Let `f(n)=sum_(k=-n)^(n)(cot^(-1)((1)/(k))-tan^(-1)(k))` such that `sum_(n=2)^(10)(f(n)+f(n-1))=a pi` then find the value of `(a+1)`.

Answer

To solve the problem, we need to evaluate the function \( f(n) = \sum_{k=-n}^{n} \left( \cot^{-1}\left(\frac{1}{k}\right) - \tan^{-1}(k) \right) \) and then compute the sum \( \sum_{n=2}^{10} (f(n) + f(n-1)) = a \pi \). Finally, we will find the value of \( a + 1 \).

### Step 1: Simplifying \( f(n) \)

We start with the function:
\[ f(n) = \sum_{k=-n}^{n} \left( \cot^{-1}\left(\frac{1}{k}\right) - \tan^{-1}(k) \right) \]

Using the identity \( \cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) \), we can rewrite \( \cot^{-1}\left(\frac{1}{k}\right) \) as \( \tan^{-1}(k) \). Thus, we have:
\[ f(n) = \sum_{k=-n}^{n} \left( \tan^{-1}(k) - \tan^{-1}(k) \right) = \sum_{k=-n}^{n} 0 = 0 \]

However, we need to be careful with \( k = 0 \) since \( \cot^{-1}(0) \) is undefined. Thus, we separate the sum into two parts:
\[ f(n) = \sum_{k=-n}^{-1} \left( \cot^{-1}\left(\frac{1}{k}\right) - \tan^{-1}(k) \right) + \sum_{k=1}^{n} \left( \cot^{-1}\left(\frac{1}{k}\right) - \tan^{-1}(k) \right) \]

### Step 2: Evaluating Each Part

For \( k < 0 \):
\[ \cot^{-1}\left(\frac{1}{k}\right) = \tan^{-1}(-k) + \pi \]
Thus,
\[ \cot^{-1}\left(\frac{1}{k}\right) - \tan^{-1}(k) = \tan^{-1}(-k) + \pi - \tan^{-1}(k) \]

For \( k > 0 \):
\[ \cot^{-1}\left(\frac{1}{k}\right) = \tan^{-1}(k) \]
Thus,
\[ \cot^{-1}\left(\frac{1}{k}\right) - \tan^{-1}(k) = 0 \]

### Step 3: Combining Results

Now we combine both parts:
\[ f(n) = \sum_{k=-n}^{-1} \left( \tan^{-1}(-k) + \pi - \tan^{-1}(k) \right) \]
Notice that \( \tan^{-1}(-k) = -\tan^{-1}(k) \):
\[ f(n) = \sum_{k=1}^{n} \left( -\tan^{-1}(k) + \pi - \tan^{-1}(k) \right) = \sum_{k=1}^{n} \left( \pi - 2\tan^{-1}(k) \right) \]
This simplifies to:
\[ f(n) = n\pi - 2\sum_{k=1}^{n} \tan^{-1}(k) \]

### Step 4: Evaluating the Final Sum

Now we need to evaluate:
\[ \sum_{n=2}^{10} (f(n) + f(n-1)) \]
Substituting \( f(n) = n\pi - 2\sum_{k=1}^{n} \tan^{-1}(k) \) and \( f(n-1) = (n-1)\pi - 2\sum_{k=1}^{n-1} \tan^{-1}(k) \):
\[ f(n) + f(n-1) = n\pi - 2\sum_{k=1}^{n} \tan^{-1}(k) + (n-1)\pi - 2\sum_{k=1}^{n-1} \tan^{-1}(k) \]
This simplifies to:
\[ (2n - 1)\pi - 2\tan^{-1}(n) \]

### Step 5: Final Calculation

Now we compute:
\[ \sum_{n=2}^{10} \left( (2n - 1)\pi - 2\tan^{-1}(n) \right) \]
This results in:
\[ \sum_{n=2}^{10} (2n - 1)\pi - 2\sum_{n=2}^{10} \tan^{-1}(n) \]
Calculating \( \sum_{n=2}^{10} (2n - 1) = 3 + 5 + 7 + 9 + 11 + 13 + 15 + 17 + 19 = 99 \).

Thus, we have:
\[ \sum_{n=2}^{10} (f(n) + f(n-1)) = 99\pi \]
So, \( a = 99 \).

### Step 6: Final Result

Finally, we need to find \( a + 1 \):
\[ a + 1 = 99 + 1 = 100 \]

Therefore, the final answer is:
\[ \boxed{100} \]