If ` a + b = 1, c + d = 1 and a - b = (d)/(c)` ,t hen the value of ` c^(2) - d^(2)` is

Options

A. `(a)/(b)`
B. `(b)/(a)`
C. 1
D. `-1`

Answer

To solve the problem step by step, we will use the given equations and manipulate them to find the value of \( c^2 - d^2 \).

### Step 1: Write down the given equations
We have the following equations:
1. \( a + b = 1 \) (Equation 1)
2. \( c + d = 1 \) (Equation 2)
3. \( a - b = \frac{d}{c} \) (Equation 3)

### Step 2: Express \( c^2 - d^2 \)
We can use the identity \( c^2 - d^2 = (c - d)(c + d) \). From Equation 2, we know that \( c + d = 1 \).

### Step 3: Find \( c - d \)
To find \( c - d \), we can manipulate Equations 1 and 3.

First, let's add Equations 1 and 3:
\[ (a + b) + (a - b) = 1 + \frac{d}{c} \]
This simplifies to:
\[ 2a = 1 + \frac{d}{c} \quad \text{(Equation 4)} \]

Next, let's subtract Equation 3 from Equation 1:
\[ (a + b) - (a - b) = 1 - \frac{d}{c} \]
This simplifies to:
\[ 2b = 1 - \frac{d}{c} \quad \text{(Equation 5)} \]

### Step 4: Solve for \( c \) and \( d \)
From Equation 4, we can express \( d \):
\[ \frac{d}{c} = 2a - 1 \implies d = c(2a - 1) \quad \text{(Equation 6)} \]

Substituting Equation 6 into Equation 2:
\[ c + c(2a - 1) = 1 \]
Factoring out \( c \):
\[ c(1 + 2a - 1) = 1 \implies c(2a) = 1 \implies c = \frac{1}{2a} \quad \text{(Equation 7)} \]

### Step 5: Substitute \( c \) back to find \( d \)
Using Equation 6:
\[ d = c(2a - 1) = \frac{1}{2a}(2a - 1) = \frac{2a - 1}{2a} \quad \text{(Equation 8)} \]

### Step 6: Calculate \( c - d \)
Now we can find \( c - d \):
\[ c - d = \frac{1}{2a} - \frac{2a - 1}{2a} = \frac{1 - (2a - 1)}{2a} = \frac{2 - 2a}{2a} = \frac{2(1 - a)}{2a} = \frac{1 - a}{a} \]

### Step 7: Substitute into \( c^2 - d^2 \)
Now substituting back into \( c^2 - d^2 \):
\[ c^2 - d^2 = (c - d)(c + d) = \left(\frac{1 - a}{a}\right)(1) = \frac{1 - a}{a} \]

### Step 8: Express in terms of \( b \)
Since \( b = 1 - a \) from Equation 1, we can substitute:
\[ c^2 - d^2 = \frac{b}{a} \]

### Final Answer
Thus, the value of \( c^2 - d^2 \) is:
\[ \frac{b}{a} \]