If `cot^(-1) 2 + cot^(-1) 3 + cot^(-1) 4 + …….+ cot^(-1) n = pi/4`then value of n is

Options

A. 2
B. 4
C. 3
D. 5

Answer

To solve the problem, we need to find the value of \( n \) such that:

\[ \cot^{-1}(2) + \cot^{-1}(3) + \cot^{-1}(4) + \ldots + \cot^{-1}(n) = \frac{\pi}{4} \]

### Step 1: Understanding the cotangent inverse addition formula
We will use the identity for the sum of two cotangent inverses:

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

This means we can combine the terms in the series using this identity.

### Step 2: Start with the first two terms
Let’s first consider the sum of the first two terms, \( \cot^{-1}(2) + \cot^{-1}(3) \):

Using the identity:

\[ \cot^{-1}(2) + \cot^{-1}(3) = \cot^{-1}\left(\frac{2 \cdot 3 - 1}{2 + 3}\right) = \cot^{-1}\left(\frac{6 - 1}{5}\right) = \cot^{-1}\left(\frac{5}{5}\right) = \cot^{-1}(1) \]

### Step 3: Recognizing the value of cotangent inverse
We know that:

\[ \cot^{-1}(1) = \frac{\pi}{4} \]

### Step 4: Conclusion about \( n \)
Since \( \cot^{-1}(2) + \cot^{-1}(3) = \frac{\pi}{4} \), we can conclude that the sum of the series from \( \cot^{-1}(2) \) to \( \cot^{-1}(n) \) must equal \( \frac{\pi}{4} \) when \( n = 3 \).

If we include any term \( \cot^{-1}(k) \) for \( k > 3 \), the sum will exceed \( \frac{\pi}{4} \).

Thus, the value of \( n \) is:

\[ \boxed{3} \]