If `0 lt x lt pi/2 and sec x = cosec y`, then the value of `sin (x + y)` is

Options

A. 0
B. 1
C. `1/2`
D. `1/sqrt3`

Answer

To solve the problem, we need to find the value of \( \sin(x + y) \) given that \( \sec x = \csc y \) and \( 0 < x < \frac{\pi}{2} \).

### Step-by-Step Solution:

1. **Start with the given equation:**
\[ \sec x = \csc y \]
Recall that \( \sec x = \frac{1}{\cos x} \) and \( \csc y = \frac{1}{\sin y} \). Thus, we can rewrite the equation as:
\[ \frac{1}{\cos x} = \frac{1}{\sin y} \]

2. **Cross-multiply to find a relationship between \( \sin y \) and \( \cos x \):**
\[ \sin y = \cos x \]

3. **Use the sine addition formula:**
The formula for \( \sin(x + y) \) is:
\[ \sin(x + y) = \sin x \cos y + \cos x \sin y \]

4. **Substitute \( \sin y \) into the formula:**
From step 2, we know \( \sin y = \cos x \). Therefore, we can substitute this into the sine addition formula:
\[ \sin(x + y) = \sin x \cos y + \cos x \cos x \]
Simplifying gives:
\[ \sin(x + y) = \sin x \cos y + \cos^2 x \]

5. **Find \( \cos y \):**
We can find \( \cos y \) using the Pythagorean identity:
\[ \cos^2 y + \sin^2 y = 1 \]
Substituting \( \sin y = \cos x \):
\[ \cos^2 y + \cos^2 x = 1 \]
Therefore,
\[ \cos^2 y = 1 - \cos^2 x \]
This simplifies to:
\[ \cos^2 y = \sin^2 x \]
Taking the square root (since \( y \) is in the first quadrant, \( \cos y \) is positive):
\[ \cos y = \sin x \]

6. **Substitute \( \cos y \) back into the sine addition formula:**
Now substitute \( \cos y = \sin x \) into the equation for \( \sin(x + y) \):
\[ \sin(x + y) = \sin x \sin x + \cos^2 x \]
This simplifies to:
\[ \sin(x + y) = \sin^2 x + \cos^2 x \]

7. **Use the Pythagorean identity:**
We know from the Pythagorean identity that:
\[ \sin^2 x + \cos^2 x = 1 \]

8. **Final result:**
Therefore, we conclude:
\[ \sin(x + y) = 1 \]

### Conclusion:
The value of \( \sin(x + y) \) is \( 1 \).