The value of `cotsum_(n=1)^(19)cot^(-1)(1+sum_(p=1)^(n)2p)` is equal to (a) `(21)/(19)` (b) `(19)/(21)` (c) `-(19)/(21)` (d) `-(21)/(19)`

Options

A. `19/21`
B. `21/19`
C. `22/23`
D. `23/22`

Explanation

To solve the given problem, we need to evaluate the expression:

\[ \sum_{n=1}^{19} \cot^{-1}\left(1 + \sum_{p=1}^{n} 2p\right) \]

### Step 1: Simplify the inner summation

First, we simplify the inner summation:

\[ \sum_{p=1}^{n} 2p = 2 \sum_{p=1}^{n} p = 2 \cdot \frac{n(n+1)}{2} = n(n+1) \]

This means we can rewrite the expression as:

\[ \sum_{n=1}^{19} \cot^{-1}(1 + n(n+1)) \]

### Step 2: Rewrite the argument of cotangent inverse

Now we can simplify the argument of the cotangent inverse:

\[ 1 + n(n+1) = n^2 + n + 1 \]

Thus, our expression becomes:

\[ \sum_{n=1}^{19} \cot^{-1}(n^2 + n + 1) \]

### Step 3: Use the cotangent difference identity

We know the identity:

\[ \cot^{-1}(x) - \cot^{-1}(y) = \cot^{-1}\left(\frac{y - x}{1 + xy}\right) \]

We can express \(\cot^{-1}(n^2 + n + 1)\) in terms of a difference of cotangent inverses. We can observe that:

\[ \cot^{-1}(n^2 + n + 1) = \tan^{-1}\left(\frac{1}{n^2 + n + 1}\right) \]

### Step 4: Evaluate the sum

Now we can evaluate the sum:

\[ \sum_{n=1}^{19} \left(\tan^{-1}(n+1) - \tan^{-1}(n)\right) \]

This is a telescoping series, where most terms will cancel out:

\[ = \tan^{-1}(20) - \tan^{-1}(1) \]

### Step 5: Use the tangent difference identity

Using the identity for the difference of arctangents:

\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a - b}{1 + ab}\right) \]

We can substitute \(a = 20\) and \(b = 1\):

\[ \tan^{-1}(20) - \tan^{-1}(1) = \tan^{-1}\left(\frac{20 - 1}{1 + 20 \cdot 1}\right) = \tan^{-1}\left(\frac{19}{21}\right) \]

### Step 6: Find cotangent of the result

Now we take the cotangent of the result:

\[ \cot\left(\tan^{-1}\left(\frac{19}{21}\right)\right) = \frac{21}{19} \]

### Final Answer

Thus, the value of the original expression is:

\[ \frac{21}{19} \]

So the correct option is **(a) \(\frac{21}{19}\)**.

---