Let `A =[-1,1]`. Define a relation R on A as follows:
`a, b in A, aRb` if and only if `sin^(-1)(a) + cos^(-1)(b) = pi//2`, Then

Options

A. R is reflexive and symmetric only
B. R is symmetric and transitive only
C. R is transitive only
D. R is an equivalence relation

Answer

To determine the properties of the relation \( R \) defined on the set \( A = [-1, 1] \) by the condition \( aRb \) if and only if \( \sin^{-1}(a) + \cos^{-1}(b) = \frac{\pi}{2} \), we will check if the relation is reflexive, symmetric, and transitive.

### Step 1: Check Reflexivity
A relation \( R \) is reflexive if every element \( a \in A \) satisfies \( aRa \).

For reflexivity:
- We need to check if \( \sin^{-1}(a) + \cos^{-1}(a) = \frac{\pi}{2} \) for all \( a \in A \).
- We know that \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \) for all \( x \) in the domain of \( \sin^{-1} \) and \( \cos^{-1} \).
- Since \( a \) is in the interval \([-1, 1]\), this condition holds.

Thus, \( R \) is reflexive.

### Step 2: Check Symmetry
A relation \( R \) is symmetric if whenever \( aRb \), then \( bRa \).

For symmetry:
- Assume \( aRb \) holds, which means \( \sin^{-1}(a) + \cos^{-1}(b) = \frac{\pi}{2} \).
- We need to check if \( bRa \) holds, i.e., \( \sin^{-1}(b) + \cos^{-1}(a) = \frac{\pi}{2} \).
- From the identity \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \), we can express:
\[ \cos^{-1}(b) = \frac{\pi}{2} - \sin^{-1}(b) \]
\[ \cos^{-1}(a) = \frac{\pi}{2} - \sin^{-1}(a) \]
- Substituting these into our original equation:
\[ \sin^{-1}(a) + \left(\frac{\pi}{2} - \sin^{-1}(b)\right) = \frac{\pi}{2} \]
Simplifying gives:
\[ \sin^{-1}(a) - \sin^{-1}(b) = 0 \implies \sin^{-1}(a) = \sin^{-1}(b) \implies a = b \]
- Thus, \( bRa \) holds.

Therefore, \( R \) is symmetric.

### Step 3: Check Transitivity
A relation \( R \) is transitive if whenever \( aRb \) and \( bRc \), then \( aRc \).

For transitivity:
- Assume \( aRb \) and \( bRc \):
- From \( aRb \): \( \sin^{-1}(a) + \cos^{-1}(b) = \frac{\pi}{2} \) (1)
- From \( bRc \): \( \sin^{-1}(b) + \cos^{-1}(c) = \frac{\pi}{2} \) (2)
- We need to show \( aRc \), which means \( \sin^{-1}(a) + \cos^{-1}(c) = \frac{\pi}{2} \).
- From (1), we can express \( \cos^{-1}(b) = \frac{\pi}{2} - \sin^{-1}(a) \).
- From (2), we can express \( \sin^{-1}(b) = \frac{\pi}{2} - \cos^{-1}(c) \).
- Substituting \( \sin^{-1}(b) \) from (2) into (1):
\[ \sin^{-1}(a) + \left(\frac{\pi}{2} - \sin^{-1}(b)\right) = \frac{\pi}{2} \]
This simplifies to:
\[ \sin^{-1}(a) + \cos^{-1}(c) = \frac{\pi}{2} \]
- Hence, \( aRc \) holds.

Thus, \( R \) is transitive.

### Conclusion
Since the relation \( R \) is reflexive, symmetric, and transitive, it is an equivalence relation.

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