In the ambiguous case, if `a, b and A` are given and `c_1, c_2` are the two values of the third `(c_1-c_2)^2 + (c_1+c_2)^2 tan^2 A` is equal to

Options

A. 4
B. `4a^(2)`
C. `4b^(2)`
D. `4c^(2)`

Explanation

To solve the given problem step by step, we need to derive the expression for \((c_1 - c_2)^2 + (c_1 + c_2)^2 \tan^2 A\) based on the information provided.

### Step 1: Understand the given values
We are given:
- \(a\), \(b\), and \(A\) (where \(A\) is an angle)
- \(c_1\) and \(c_2\) are the two possible values of the side \(c\) in the ambiguous case.

### Step 2: Use the properties of roots
From the quadratic equation formed by the sides \(a\), \(b\), and angle \(A\), we know:
- The sum of the roots \(c_1 + c_2 = 2b \cos A\)
- The product of the roots \(c_1 c_2 = b^2 - a^2\)

### Step 3: Calculate \(c_1 - c_2\) and \(c_1 + c_2\)
Using the sum and product of roots:
- \(c_1 - c_2 = \sqrt{(c_1 + c_2)^2 - 4c_1 c_2}\)
- Substitute the values:
\[ c_1 - c_2 = \sqrt{(2b \cos A)^2 - 4(b^2 - a^2)} \]

### Step 4: Simplify \(c_1 - c_2\)
Calculating:
\[ (2b \cos A)^2 = 4b^2 \cos^2 A \]
Thus:
\[ c_1 - c_2 = \sqrt{4b^2 \cos^2 A - 4(b^2 - a^2)} = \sqrt{4b^2 \cos^2 A - 4b^2 + 4a^2} \]
This simplifies to:
\[ c_1 - c_2 = \sqrt{4a^2 - 4b^2(1 - \cos^2 A)} = \sqrt{4a^2 - 4b^2 \sin^2 A} \]

### Step 5: Calculate \((c_1 - c_2)^2\)
Squaring the expression:
\[ (c_1 - c_2)^2 = 4a^2 - 4b^2 \sin^2 A \]

### Step 6: Calculate \((c_1 + c_2)^2\)
Now calculate \((c_1 + c_2)^2\):
\[ (c_1 + c_2)^2 = (2b \cos A)^2 = 4b^2 \cos^2 A \]

### Step 7: Combine the results
Now we need to evaluate:
\[ (c_1 - c_2)^2 + (c_1 + c_2)^2 \tan^2 A \]
Substituting the values:
\[ = (4a^2 - 4b^2 \sin^2 A) + (4b^2 \cos^2 A) \tan^2 A \]
Since \(\tan^2 A = \frac{\sin^2 A}{\cos^2 A}\):
\[ = 4a^2 - 4b^2 \sin^2 A + 4b^2 \sin^2 A = 4a^2 \]

### Final Answer
Thus, we conclude that:
\[ (c_1 - c_2)^2 + (c_1 + c_2)^2 \tan^2 A = 4a^2 \]