.Let `a_(1)=1,a_(2),a_(3),a_(4),....` be consecutive natural number,then `tan^(-1)((1)/(1+a_(1)a_(2)))+tan^(-1)((1)/(1+a_(2)a_(3)))+ . . . +tan^(-1)((1)/(1+a_(2021)a_(2022)))` is equal to

Options

A. `cot^(-1)(2022)-(pi)/(4)`
B. `(pi)/(4)-cot^(-1)(2022)`
C. `tan^(-1)(2022)-(pi)/(4)`
D. `(pi)/(4)-tan^(-1)(2022)`

Answer

To solve the problem, we need to evaluate the sum:

\[ S = \tan^{-1}\left(\frac{1}{1 + a_1 a_2}\right) + \tan^{-1}\left(\frac{1}{1 + a_2 a_3}\right) + \ldots + \tan^{-1}\left(\frac{1}{1 + a_{2021} a_{2022}}\right) \]

where \( a_n \) are consecutive natural numbers starting from \( a_1 = 1 \).

### Step 1: Identify the terms
The terms can be expressed as:

- \( a_1 = 1 \)
- \( a_2 = 2 \)
- \( a_3 = 3 \)
- ...
- \( a_{2022} = 2022 \)

### Step 2: Rewrite the terms using the formula
Using the identity \( \tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\left(\frac{x+y}{1-xy}\right) \) when \( xy < 1 \), we can rewrite each term:

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

### Step 3: Apply the identity
Thus, we can rewrite the sum \( S \):

\[ S = \left( \tan^{-1}(2 - 1) - \tan^{-1}(1) \right) + \left( \tan^{-1}(3 - 2) - \tan^{-1}(2) \right) + \ldots + \left( \tan^{-1}(2022 - 2021) - \tan^{-1}(2021) \right) \]

### Step 4: Simplify the expression
Notice that this is a telescoping series:

\[ S = \tan^{-1}(2022 - 1) - \tan^{-1}(1) = \tan^{-1}(2021) - \tan^{-1}(1) \]

### Step 5: Evaluate the final expression
We know that \( \tan^{-1}(1) = \frac{\pi}{4} \). Thus, we have:

\[ S = \tan^{-1}(2021) - \frac{\pi}{4} \]

### Step 6: Final answer
The final answer for the sum is:

\[ S = \tan^{-1}(2021) - \frac{\pi}{4} \]