Consider the following statements :
1. `tan^(-1) x + tan^(-1)((1)/(x))=pi`
2. There exist x, y `in[-1, 1]`, where x `ne` y such that `sin^(-1) x + cos^(-1) y = (pi)/(2)`.
Which of the above statement is/are correct ?

Options

A. 1 only
B. 2 only
C. Both 1 and 2
D. Neither 1 nor 2

Explanation

To solve the problem, we need to evaluate the two statements given and determine their correctness.

### Statement 1:
**tan^(-1) x + tan^(-1)(1/x) = π**

To analyze this statement, we can use the identity for the sum of inverse tangents:
\[ \tan^{-1} a + \tan^{-1} b = \tan^{-1} \left( \frac{a + b}{1 - ab} \right) \]
when \( ab < 1 \).

In our case, let \( a = x \) and \( b = \frac{1}{x} \). Then:
\[ \tan^{-1} x + \tan^{-1} \left( \frac{1}{x} \right) = \tan^{-1} \left( \frac{x + \frac{1}{x}}{1 - x \cdot \frac{1}{x}} \right) = \tan^{-1} \left( \frac{x + \frac{1}{x}}{0} \right) \]
The denominator becomes zero, which means the expression is undefined, and we approach infinity.

However, we know that:
\[ \tan^{-1} x + \tan^{-1} \left( \frac{1}{x} \right) = \frac{\pi}{2} \quad \text{for } x > 0 \]
and
\[ \tan^{-1} x + \tan^{-1} \left( \frac{1}{x} \right) = -\frac{\pi}{2} \quad \text{for } x < 0 \]
Thus, the first statement is incorrect because it claims the sum equals π instead of \(\frac{\pi}{2}\).

### Statement 2:
**There exist x, y ∈ [-1, 1], where x ≠ y such that sin^(-1) x + cos^(-1) y = π/2.**

Using the identity:
\[ \sin^{-1} x + \cos^{-1} x = \frac{\pi}{2} \]
This holds true when \( x = y \). However, the statement specifies that \( x \neq y \).

If we let \( y = \sqrt{1 - x^2} \) (which is valid for \( x \in [-1, 1] \)), then:
\[ \sin^{-1} x + \cos^{-1} \sqrt{1 - x^2} = \frac{\pi}{2} \]
But since \( y \) must equal \( x \) for the identity to hold, and we are given \( x \neq y \), this statement is also incorrect.

### Conclusion:
Both statements are incorrect.

### Final Answer:
Neither statement 1 nor statement 2 is correct.

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