If `ax + b sec(tan^-1 x) = c and ay + b sec(tan^-ly) = c,` then `(x+y)/(1-xy)` is equal to

Options

A. `(2ab)/(a^(2) -b^(2))`
B. `(c^(2) -b^(2))/(a^(2) -b^(2))`
C. `(c^(2) -b^(2))/(a^(2) + b^(2))`
D. none of these

Explanation

To solve the problem step by step, we start with the given equations:

1. **Given Equations:**
\[ ax + b \sec(\tan^{-1} x) = c \]
\[ ay + b \sec(\tan^{-1} y) = c \]

2. **Substituting Variables:**
Let \( \tan^{-1} x = \alpha \) and \( \tan^{-1} y = \beta \). Therefore, we have:
\[ x = \tan(\alpha) \quad \text{and} \quad y = \tan(\beta) \]

3. **Rewriting the Equations:**
The equations become:
\[ a \tan(\alpha) + b \sec(\alpha) = c \]
\[ a \tan(\beta) + b \sec(\beta) = c \]

4. **Isolating Secant:**
From both equations, we can isolate \( b \sec(\alpha) \) and \( b \sec(\beta) \):
\[ b \sec(\alpha) = c - a \tan(\alpha) \]
\[ b \sec(\beta) = c - a \tan(\beta) \]

5. **Squaring Both Sides:**
Squaring both equations gives:
\[ b^2 \sec^2(\alpha) = (c - a \tan(\alpha))^2 \]
\[ b^2 \sec^2(\beta) = (c - a \tan(\beta))^2 \]

6. **Using the Identity:**
Recall that \( \sec^2(\theta) = 1 + \tan^2(\theta) \):
\[ b^2 (1 + \tan^2(\alpha)) = (c - a \tan(\alpha))^2 \]
\[ b^2 (1 + \tan^2(\beta)) = (c - a \tan(\beta))^2 \]

7. **Setting Up a Quadratic Equation:**
This leads us to a quadratic equation in terms of \( \tan(\alpha) \) and \( \tan(\beta) \). The sum of the roots \( \tan(\alpha) + \tan(\beta) \) can be expressed as:
\[ \tan(\alpha) + \tan(\beta) = \frac{c^2 - b^2}{a^2 - b^2} \]

8. **Finding \( \frac{x+y}{1-xy} \):**
Using the identity for the tangent of the sum of angles:
\[ \tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha) \tan(\beta)} \]
We can express:
\[ \frac{x+y}{1-xy} = \tan(\alpha + \beta) \]

9. **Final Expression:**
Hence, we find:
\[ \frac{x+y}{1-xy} = \frac{2ac}{a^2 - c^2} \]

Thus, the final answer is:
\[ \frac{x+y}{1-xy} = \frac{2ac}{a^2 - c^2} \]