Statement 1: The value of `cot^(-1)((7)/(4))+cot^(-1)((19)/(4))+...+cot^(-1)((4r^(2)+3)/(4))+...` upto infinity is `tan^(-1)(2)`.
Statement 2: `sum_(r=1)^(n)"tan"(x_(r)-x_(r-1))/(1+x_(r-1)x_(r))=tan^(-1)x_(n)-tan^(-1)x_(0)`.

Options

A. Statement-1 is True, Statement-2 is True and Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True and Statement-2 is NOT a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Answer

To solve the problem, we will analyze both statements step by step.

### Statement 1:
We need to evaluate the infinite series:

\[ S = \cot^{-1}\left(\frac{7}{4}\right) + \cot^{-1}\left(\frac{19}{4}\right) + \ldots + \cot^{-1}\left(\frac{4r^2 + 3}{4}\right) + \ldots \]

1. **Identify the general term**: The general term can be expressed as:
\[ T_r = \cot^{-1}\left(\frac{4r^2 + 3}{4}\right) \]

2. **Convert cotangent inverse to tangent inverse**: We know that:
\[ \cot^{-1}(x) = \tan^{-1}\left(\frac{1}{x}\right) \]
Thus,
\[ T_r = \tan^{-1}\left(\frac{4}{4r^2 + 3}\right) \]

3. **Rewrite the term**: We can express \(T_r\) in a more manageable form:
\[ T_r = \tan^{-1}\left(\frac{4}{4r^2 + 3}\right) = \tan^{-1}\left(\frac{4}{4(r^2 + \frac{3}{4})}\right) = \tan^{-1}\left(\frac{1}{r^2 + \frac{3}{4}}\right) \]

4. **Use the tangent addition formula**: The series can be summed up using the formula:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a - b}{1 + ab}\right) \]
By recognizing that \(T_r\) can be expressed as:
\[ T_r = \tan^{-1}\left(r + \frac{1}{2}\right) - \tan^{-1}\left(r - \frac{1}{2}\right) \]

5. **Sum the series**: The series telescopes:
\[ S = \sum_{r=1}^{\infty} \left(\tan^{-1}\left(r + \frac{1}{2}\right) - \tan^{-1}\left(r - \frac{1}{2}\right)\right) \]
Most terms cancel out, leading to:
\[ S = \lim_{n \to \infty} \left(\tan^{-1}\left(n + \frac{1}{2}\right) - \tan^{-1}\left(\frac{1}{2}\right)\right) \]

6. **Evaluate the limit**: As \(n\) approaches infinity, \(\tan^{-1}(n + \frac{1}{2})\) approaches \(\frac{\pi}{2}\):
\[ S = \frac{\pi}{2} - \tan^{-1}\left(\frac{1}{2}\right) \]

7. **Convert to cotangent form**: We know that:
\[ \tan^{-1}\left(\frac{1}{2}\right) = \cot^{-1}(2) \]
Thus,
\[ S = \tan^{-1}(2) \]

### Conclusion for Statement 1:
The value of the infinite series is indeed \(\tan^{-1}(2)\), so Statement 1 is true.

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### Statement 2:
We need to verify the identity:

\[ \sum_{r=1}^{n} \tan^{-1}\left(\frac{x_r - x_{r-1}}{1 + x_{r-1}x_r}\right) = \tan^{-1}(x_n) - \tan^{-1}(x_0) \]

1. **Recognize the left-hand side**: The left-hand side can be rewritten using the property of the tangent inverse:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a - b}{1 + ab}\right) \]

2. **Apply the formula**: If we denote \(x_r\) and \(x_{r-1}\) as two consecutive terms, we can express:
\[ \sum_{r=1}^{n} \tan^{-1}\left(\frac{x_r - x_{r-1}}{1 + x_{r-1}x_r}\right) = \tan^{-1}(x_n) - \tan^{-1}(x_0) \]

3. **Telescoping nature**: The series telescopes, meaning that all intermediate terms cancel out, leading to:
\[ \tan^{-1}(x_n) - \tan^{-1}(x_0) \]

### Conclusion for Statement 2:
The identity holds true, so Statement 2 is also true.

### Final Conclusion:
Both statements are true, and Statement 2 provides a valid explanation for Statement 1.

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