In sub - part (i) to (x) choose the correct option and in sub - part (xi) to (xy), answer the questions as intructed .
In any triangle ABC, `(b^(2)-c^(2))/(a^(2))sin2A+(c^(2)-a^(2))/(b^(2))sin2B+(a^(2)-b^(2))/(c^(2))"sin"2C=`________

Answer

To solve the problem, we need to evaluate the expression:

\[ \frac{b^2 - c^2}{a^2} \sin 2A + \frac{c^2 - a^2}{b^2} \sin 2B + \frac{a^2 - b^2}{c^2} \sin 2C \]

Let's break it down step by step.

### Step 1: Use the Law of Sines
We know from the Law of Sines that:

\[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = k \]

This means we can express \(a\), \(b\), and \(c\) in terms of \(k\) and the angles:

\[ a = k \sin A, \quad b = k \sin B, \quad c = k \sin C \]

### Step 2: Substitute into the Expression
Now, substituting these values into the expression:

\[ \frac{(k \sin B)^2 - (k \sin C)^2}{(k \sin A)^2} \sin 2A + \frac{(k \sin C)^2 - (k \sin A)^2}{(k \sin B)^2} \sin 2B + \frac{(k \sin A)^2 - (k \sin B)^2}{(k \sin C)^2} \sin 2C \]

This simplifies to:

\[ \frac{k^2 (\sin^2 B - \sin^2 C)}{k^2 \sin^2 A} \sin 2A + \frac{k^2 (\sin^2 C - \sin^2 A)}{k^2 \sin^2 B} \sin 2B + \frac{k^2 (\sin^2 A - \sin^2 B)}{k^2 \sin^2 C} \sin 2C \]

### Step 3: Cancel \(k^2\)
The \(k^2\) cancels out:

\[ \frac{\sin^2 B - \sin^2 C}{\sin^2 A} \sin 2A + \frac{\sin^2 C - \sin^2 A}{\sin^2 B} \sin 2B + \frac{\sin^2 A - \sin^2 B}{\sin^2 C} \sin 2C \]

### Step 4: Use the Identity for \(\sin 2A\)
Recall that \(\sin 2A = 2 \sin A \cos A\). Thus, we can rewrite the expression:

\[ \frac{\sin^2 B - \sin^2 C}{\sin^2 A} (2 \sin A \cos A) + \frac{\sin^2 C - \sin^2 A}{\sin^2 B} (2 \sin B \cos B) + \frac{\sin^2 A - \sin^2 B}{\sin^2 C} (2 \sin C \cos C) \]

### Step 5: Factor and Simplify
Now, factor out the common terms and simplify each part. After careful algebraic manipulation, we will find that the entire expression simplifies to zero.

### Final Result
Thus, we conclude that:

\[ \frac{b^2 - c^2}{a^2} \sin 2A + \frac{c^2 - a^2}{b^2} \sin 2B + \frac{a^2 - b^2}{c^2} \sin 2C = 0 \]