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

Explanation

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

1. \( ax + b \sec(\tan^{-1} x) = c \)
2. \( ay + b \sec(\tan^{-1} y) = c \)

### Step 1: Substitute Variables
Let \( \tan^{-1} x = \alpha \) and \( \tan^{-1} y = \beta \). Then we can rewrite the equations as:

\[ a \tan \alpha + b \sec \alpha = c \]
\[ a \tan \beta + b \sec \beta = c \]

### Step 2: Rearranging the Equations
From both equations, we can express \( b \sec \alpha \) and \( b \sec \beta \):

\[ b \sec \alpha = c - a \tan \alpha \]
\[ b \sec \beta = c - a \tan \beta \]

### Step 3: Square Both Sides
Now, square both sides of the equations:

\[ b^2 \sec^2 \alpha = (c - a \tan \alpha)^2 \]
\[ b^2 \sec^2 \beta = (c - a \tan \beta)^2 \]

### Step 4: Use the Identity for Secant
Recall that \( \sec^2 \theta = 1 + \tan^2 \theta \). Thus, we can rewrite the equations as:

\[ b^2 (1 + \tan^2 \alpha) = c^2 - 2ac \tan \alpha + a^2 \tan^2 \alpha \]
\[ b^2 (1 + \tan^2 \beta) = c^2 - 2ac \tan \beta + a^2 \tan^2 \beta \]

### Step 5: Combine the Equations
Now, we can combine the equations:

\[ b^2 + b^2 \tan^2 \alpha = c^2 - 2ac \tan \alpha + a^2 \tan^2 \alpha \]
\[ b^2 + b^2 \tan^2 \beta = c^2 - 2ac \tan \beta + a^2 \tan^2 \beta \]

### Step 6: Rearranging and Collecting Terms
Rearranging gives us:

\[ (b^2 - a^2) \tan^2 \alpha + 2ac \tan \alpha + (b^2 - c^2) = 0 \]
\[ (b^2 - a^2) \tan^2 \beta + 2ac \tan \beta + (b^2 - c^2) = 0 \]

### Step 7: Sum and Product of Roots
From the quadratic equations, we can identify the sum and product of the roots:

- Sum of roots \( \tan \alpha + \tan \beta = -\frac{2ac}{b^2 - a^2} \)
- Product of roots \( \tan \alpha \tan \beta = \frac{b^2 - c^2}{b^2 - a^2} \)

### Step 8: Finding \( \frac{x+y}{1-xy} \)
We know that:

\[ \tan(\alpha + \beta) = \frac{\tan \alpha + \tan \beta}{1 - \tan \alpha \tan \beta} \]

Substituting the values we found:

\[ \frac{x + y}{1 - xy} = \frac{-\frac{2ac}{b^2 - a^2}}{1 - \frac{b^2 - c^2}{b^2 - a^2}} \]

### Step 9: Simplifying the Expression
Now simplify the denominator:

\[ 1 - \frac{b^2 - c^2}{b^2 - a^2} = \frac{(b^2 - a^2) - (b^2 - c^2)}{b^2 - a^2} = \frac{c^2 - a^2}{b^2 - a^2} \]

Thus, we have:

\[ \frac{x + y}{1 - xy} = \frac{-\frac{2ac}{b^2 - a^2}}{\frac{c^2 - a^2}{b^2 - a^2}} = -\frac{2ac}{c^2 - a^2} \]

### Final Result
Therefore, the value of \( \frac{x+y}{1-xy} \) is:

\[ \frac{2ac}{a^2 - c^2} \]