When any two sides and one of the opposite acute angle are given, under certain additional conditions two triangles are possible. The case when two triangles are possible is called the ambiguous case.
In fact when any two sides and the angle opposite to one of them are given either no triangle is posible or only one triangle is possible or two triangles are possible.
In the ambiguous case, let a,b and `angle A` are given and `c_(1), c_(2)` are two values of the third side c.
On the basis of above information, answer the following questions
The value of `c_(1)^(2) -2c_(1) c_(2) cos 2A +c_(2)^(2)` is

Options

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

Answer

To solve the problem, we need to evaluate the expression \( c_1^2 - 2c_1c_2 \cos 2A + c_2^2 \).

### Step 1: Rewrite the expression
We start with the expression:
\[ c_1^2 - 2c_1c_2 \cos 2A + c_2^2 \]
This can be recognized as a form of the square of a binomial:
\[ (c_1 - c_2 \cos 2A)^2 + c_2^2 \sin^2 2A \]

### Step 2: Use the cosine double angle identity
We know that:
\[ \cos 2A = 2\cos^2 A - 1 \]
We can also express \( \sin^2 2A \) using the identity:
\[ \sin^2 2A = 1 - \cos^2 2A \]

### Step 3: Substitute and simplify
Substituting the identities into our expression, we can simplify:
\[ c_1^2 - 2c_1c_2(2\cos^2 A - 1) + c_2^2 \]
This leads us to:
\[ c_1^2 + c_2^2 - 2c_1c_2(2\cos^2 A - 1) \]

### Step 4: Recognize the quadratic form
The expression can be viewed as a quadratic in terms of \( c \):
\[ c^2 - 2bc \cos A + b^2 - a^2 = 0 \]
where \( b = c_2 \) and \( a = c_1 \).

### Step 5: Apply the quadratic formula
Using the quadratic formula, we find that the roots \( c_1 \) and \( c_2 \) satisfy:
\[ c_1 + c_2 = 2b \cos A \]
and
\[ c_1c_2 = b^2 - a^2 \]

### Step 6: Substitute back into the expression
Now we can substitute these values back into our expression:
\[ c_1^2 + c_2^2 = (c_1 + c_2)^2 - 2c_1c_2 \]
This gives us:
\[ (2b \cos A)^2 - 2(b^2 - a^2) \]
Simplifying this yields:
\[ 4b^2 \cos^2 A - 2b^2 + 2a^2 \]

### Step 7: Final simplification
Factoring out common terms, we get:
\[ 2(2b^2 \cos^2 A + a^2 - b^2) \]
This simplifies to:
\[ 4a^2 \cos^2 A \]

### Conclusion
Thus, the final value of the expression \( c_1^2 - 2c_1c_2 \cos 2A + c_2^2 \) is:
\[ \boxed{4a^2 \cos^2 A} \]