Analyzing Inverse Trigonometric Statements

Let's carefully examine each statement involving inverse trigonometric functions to determine its correctness.

Statement 1 Analysis: \({\tan ^{ - 1}}{\rm{x}} + {\tan ^{ - 1}}\left( {\frac{1}{{\rm{x}}}} \right) = {\rm{\pi }}\)

This statement involves the sum of inverse tangent functions. The relationship between \({\tan ^{ - 1}}{\rm{x}}\) and \({\tan ^{ - 1}}\left( {\frac{1}{{\rm{x}}}} \right)\) depends on the sign of x.

• Case 1: x > 0
  For x > 0, we know that \({\tan ^{ - 1}}\left( {\frac{1}{{\rm{x}}}} \right) = {\cot ^{ - 1}}{\rm{x}}\).
  We also have the identity \({\tan ^{ - 1}}{\rm{x}} + {\cot ^{ - 1}}{\rm{x}} = \frac{{\rm{\pi }}}{2}\).
  Therefore, for x > 0, \({\tan ^{ - 1}}{\rm{x}} + {\tan ^{ - 1}}\left( {\frac{1}{{\rm{x}}}} \right) = {\tan ^{ - 1}}{\rm{x}} + {\cot ^{ - 1}}{\rm{x}} = \frac{{\rm{\pi }}}{2}\).
• Case 2: x < 0
  For x < 0, the relationship is \({\tan ^{ - 1}}{\rm{x}} + {\tan ^{ - 1}}\left( {\frac{1}{{\rm{x}}}} \right) = - \frac{{\rm{\pi }}}{2}\). This can be derived using properties like \({\tan ^{ - 1}}{\rm{y}} + {\tan ^{ - 1}}{\rm{z}} = {\rm{\pi }} + {\tan ^{ - 1}}\left( {\frac{{{\rm{y}} + {\rm{z}}}}{{1 - {\rm{yz}}}}} \right)\) when y < 0, z < 0, and yz > 1, or by relating \({\tan ^{ - 1}}\left( {\frac{1}{{\rm{x}}}} \right)\) to \({\cot ^{ - 1}}{\rm{x}}\) for negative x.
  For x < 0, \({\cot ^{ - 1}}{\rm{x}} = {\rm{\pi }} + {\tan ^{ - 1}}\left( {\frac{1}{{\rm{x}}}} \right)\). Thus, \({\tan ^{ - 1}}\left( {\frac{1}{{\rm{x}}}} \right) = {\cot ^{ - 1}}{\rm{x}} - {\rm{\pi }} = \left( {\frac{{\rm{\pi }}}{2} - {\tan ^{ - 1}}{\rm{x}}} \right) - {\rm{\pi }} = - \frac{{\rm{\pi }}}{2} - {\tan ^{ - 1}}{\rm{x}}\).
  So, for x < 0, \({\tan ^{ - 1}}{\rm{x}} + {\tan ^{ - 1}}\left( {\frac{1}{{\rm{x}}}} \right) = {\tan ^{ - 1}}{\rm{x}} + \left( { - \frac{{\rm{\pi }}}{2} - {\tan ^{ - 1}}{\rm{x}}} \right) = - \frac{{\rm{\pi }}}{2}\).

In neither case (x > 0 or x < 0) does \({\tan ^{ - 1}}{\rm{x}} + {\tan ^{ - 1}}\left( {\frac{1}{{\rm{x}}}} \right)\) equal \({\rm{\pi }}\). Therefore, Statement 1 is incorrect.

Statement 2 Analysis: Existence of x, y ∈ [-1, 1], x ≠ y such that sin^-1 x + cos^-1 y = \(\frac{{\rm{\pi }}}{2}\)

This statement asks if there exist distinct values x and y within the interval [-1, 1] that satisfy the given equation involving inverse sine and inverse cosine functions.

We know a fundamental identity for inverse trigonometric functions: For any real number z in the domain [-1, 1], \({\sin ^{ - 1}}{\rm{z}} + {\cos ^{ - 1}}{\rm{z}} = \frac{{\rm{\pi }}}{2}\).

The given equation is \({\sin ^{ - 1}}{\rm{x}} + {\cos ^{ - 1}}{\rm{y}} = \frac{{\rm{\pi }}}{2}\).

Let's compare this with the known identity. If we replace z with y in the identity, we get \({\sin ^{ - 1}}{\rm{y}} + {\cos ^{ - 1}}{\rm{y}} = \frac{{\rm{\pi }}}{2}\).

So, we have:

• \({\sin ^{ - 1}}{\rm{x}} + {\cos ^{ - 1}}{\rm{y}} = \frac{{\rm{\pi }}}{2}\) (Given)
• \({\sin ^{ - 1}}{\rm{y}} + {\cos ^{ - 1}}{\rm{y}} = \frac{{\rm{\pi }}}{2}\) (Identity)

Equating the left-hand sides, we get:

\({\sin ^{ - 1}}{\rm{x}} + {\cos ^{ - 1}}{\rm{y}} = {\sin ^{ - 1}}{\rm{y}} + {\cos ^{ - 1}}{\rm{y}}\)

Subtracting \({\cos ^{ - 1}}{\rm{y}}\) from both sides:

\({\sin ^{ - 1}}{\rm{x}} = {\sin ^{ - 1}}{\rm{y}}\)

The inverse sine function, \({\sin ^{ - 1}}\), is one-to-one on its domain [-1, 1]. This means that if \({\sin ^{ - 1}}{\rm{x}} = {\sin ^{ - 1}}{\rm{y}}\) for x, y ∈ [-1, 1], then it must be the case that x = y.

Therefore, the equation \({\sin ^{ - 1}}{\rm{x}} + {\cos ^{ - 1}}{\rm{y}} = \frac{{\rm{\pi }}}{2}\) holds for x, y ∈ [-1, 1] if and only if x = y. The statement claims there exist x, y ∈ [-1, 1] where x ≠ y that satisfy this equation. This is false.

Thus, Statement 2 is incorrect.

Conclusion on Statements Correctness

Based on the analysis:

• Statement 1: \({\tan ^{ - 1}}{\rm{x}} + {\tan ^{ - 1}}\left( {\frac{1}{{\rm{x}}}} \right) = {\rm{\pi }}\) is incorrect.
• Statement 2: There exist x, y ∈ [-1, 1], where x ≠ y such that sin^-1 x + cos^-1 y = \(\frac{{\rm{\pi }}}{2}\) is incorrect.

Neither of the given statements is correct.

Revision Table: Inverse Trigonometric Identities

It's helpful to remember key identities for inverse trigonometric functions:

Additional Information on Inverse Trigonometric Functions

Inverse trigonometric functions are the inverse functions of the trigonometric functions. They are used to find the angle when the value of the trigonometric function is known. Since trigonometric functions are periodic, their inverse functions require restricting the domain to make them one-to-one.

• Domain and Range: Understanding the domain and range of each inverse trigonometric function is crucial for solving problems and evaluating expressions. For instance, \({\sin ^{ - 1}}{\rm{x}}\) has a domain of [-1, 1] and a range of \([-\frac{{\rm{\pi }}}{2}, \frac{{\rm{\pi }}}{2}]\). \({\tan ^{ - 1}}{\rm{x}}\) has a domain of all real numbers (R) and a range of \((-\frac{{\rm{\pi }}}{2}, \frac{{\rm{\pi }}}{2})\). \({\cos ^{ - 1}}{\rm{x}}\) has a domain of [-1, 1] and a range of [0, \({\rm{\pi }}\)].
• Principal Values: The range of the inverse trigonometric functions corresponds to the principal values. When solving equations or evaluating expressions, the answer must fall within this defined range.
• Identities: Various identities relate different inverse trigonometric functions, such as the reciprocal identities (\({\cot ^{ - 1}}{\rm{x}} = {\tan ^{ - 1}}\left( {\frac{1}{{\rm{x}}}} \right)\) for x > 0, etc.) and sum/difference identities. These identities are essential tools for simplifying expressions and proving relationships.

Mastering these concepts and identities is key to tackling problems involving inverse trigonometric functions.