If ` sec^(-1) ((1 + x)/(1-y)) = a , " then " (dy)/(dx) ` is

Options

A. `(y-1)/(x+1)`
B. `(y + 1)/( x-1)`
C. `(x-1)/(y-1)`
D. `(x-1)/(y+1)`

Explanation

To solve the problem, we need to find the derivative \( \frac{dy}{dx} \) given the equation:

\[ \sec^{-1} \left( \frac{1 + x}{1 - y} \right) = a \]

### Step-by-Step Solution:

1. **Rewrite the equation in terms of secant**:
\[ \frac{1 + x}{1 - y} = \sec(a) \]

2. **Cross-multiply to eliminate the fraction**:
\[ 1 + x = \sec(a) \cdot (1 - y) \]

3. **Distribute \( \sec(a) \)**:
\[ 1 + x = \sec(a) - \sec(a) y \]

4. **Rearranging the equation to isolate \( y \)**:
\[ \sec(a) y = \sec(a) - (1 + x) \]
\[ y = \frac{\sec(a) - (1 + x)}{\sec(a)} \]

5. **Differentiate both sides with respect to \( x \)**:
Using implicit differentiation:
\[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{\sec(a) - (1 + x)}{\sec(a)} \right) \]

6. **Since \( \sec(a) \) is a constant with respect to \( x \)**, we can differentiate:
\[ \frac{dy}{dx} = \frac{0 - 1}{\sec(a)} = -\frac{1}{\sec(a)} \]

7. **Substituting back for \( \sec(a) \)**:
Recall that \( \sec(a) = \frac{1 + x}{1 - y} \):
\[ \frac{dy}{dx} = -\frac{1}{\frac{1 + x}{1 - y}} = -\frac{1 - y}{1 + x} \]

8. **Final expression for \( \frac{dy}{dx} \)**:
\[ \frac{dy}{dx} = \frac{y - 1}{1 + x} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{y - 1}{1 + x} \]