Statement-1: Every function can be uniquely expressed as the sum of an even function and an odd function.
Statement-2: The set of values of parameter a for which the functions f(x) defined as ` f(x)=tan(sinx)+[(x^(2))/(a)]` on the set [-3,3] is an odd function is , `(9,oo)`

Options

A. Statement-1 is True, Statement-2 is True, statement-2 is a correct explanation for the statement-1 .
B. Statement-1 is True, Statement-2 is True, statement-2 is not a correct explanation for the statement-1 .
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False , Statement-2 is True.

Explanation

To solve the given problem, we need to analyze both statements provided.

### Step 1: Analyze Statement 1
**Statement 1**: Every function can be uniquely expressed as the sum of an even function and an odd function.

To prove this, we can take any function \( f(x) \). We can express it as follows:

1. Define the even part of the function:
\[ f_e(x) = \frac{f(x) + f(-x)}{2} \]
This is an even function because:
\[ f_e(-x) = \frac{f(-x) + f(x)}{2} = f_e(x) \]

2. Define the odd part of the function:
\[ f_o(x) = \frac{f(x) - f(-x)}{2} \]
This is an odd function because:
\[ f_o(-x) = \frac{f(-x) - f(x)}{2} = -f_o(x) \]

3. Now, we can express \( f(x) \) as:
\[ f(x) = f_e(x) + f_o(x) \]
This shows that any function can be expressed as the sum of an even function and an odd function.

**Conclusion**: Statement 1 is true.

### Step 2: Analyze Statement 2
**Statement 2**: The set of values of parameter \( a \) for which the function \( f(x) = \tan(\sin x) + \frac{x^2}{a} \) is an odd function on the interval \([-3, 3]\) is \( (9, \infty) \).

To determine when \( f(x) \) is an odd function, we need to check the condition:
\[ f(-x) = -f(x) \]

1. Calculate \( f(-x) \):
\[ f(-x) = \tan(\sin(-x)) + \frac{(-x)^2}{a} = -\tan(\sin x) + \frac{x^2}{a} \]
Since \( \sin(-x) = -\sin x \), we have:
\[ f(-x) = -\tan(\sin x) + \frac{x^2}{a} \]

2. Set the condition for \( f(x) \) to be odd:
\[ f(-x) = -f(x) \implies -\tan(\sin x) + \frac{x^2}{a} = -\left(\tan(\sin x) + \frac{x^2}{a}\right) \]
Simplifying this gives:
\[ -\tan(\sin x) + \frac{x^2}{a} = -\tan(\sin x) - \frac{x^2}{a} \]
Rearranging leads to:
\[ 2\frac{x^2}{a} = 0 \]
This implies:
\[ \frac{x^2}{a} = 0 \implies x^2 = 0 \text{ (only when } x = 0\text{)} \]

3. For \( f(x) \) to be odd for all \( x \) in \([-3, 3]\), we need:
\[ \frac{x^2}{a} < 1 \text{ for } x^2 \in [0, 9] \]
This means:
\[ a > 9 \]

**Conclusion**: Statement 2 is true.

### Final Conclusion
Both statements are true, and thus the answer is confirmed.

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