Find the value of 'a' for which the function f defined as
`f(x) = {(a sin""(pi)/(2)(x+1)",",x le 0),((tan x - sin x)/(x^(3))",", x gt 0):}` is continuous at x = 0.

Answer

To find the value of 'a' for which the function \( f \) is continuous at \( x = 0 \), we need to ensure that the left-hand limit (LHL) at \( x = 0 \) is equal to the right-hand limit (RHL) at \( x = 0 \).

The function is defined as:
\[ f(x) = \begin{cases} a \sin\left(\frac{\pi}{2}(x + 1)\right) & \text{if } x \leq 0 \\ \frac{\tan x - \sin x}{x^3} & \text{if } x > 0 \end{cases} \]

### Step 1: Calculate the Left-Hand Limit (LHL)
For \( x \leq 0 \), we will evaluate the LHL by substituting \( x = 0 \) into the first part of the function:
\[ \text{LHL} = a \sin\left(\frac{\pi}{2}(0 + 1)\right) = a \sin\left(\frac{\pi}{2}\right) = a \cdot 1 = a \]

### Step 2: Calculate the Right-Hand Limit (RHL)
For \( x > 0 \), we need to find the limit of the second part of the function as \( x \) approaches \( 0 \):
\[ \text{RHL} = \lim_{x \to 0} \frac{\tan x - \sin x}{x^3} \]

### Step 3: Evaluate the Limit
Both \( \tan x \) and \( \sin x \) approach \( 0 \) as \( x \) approaches \( 0 \), leading to the indeterminate form \( \frac{0}{0} \). We can apply L'Hôpital's Rule or simplify the expression.

Using L'Hôpital's Rule:
1. Differentiate the numerator and denominator:
- The derivative of \( \tan x \) is \( \sec^2 x \).
- The derivative of \( \sin x \) is \( \cos x \).
- The derivative of \( x^3 \) is \( 3x^2 \).

Thus, we have:
\[ \text{RHL} = \lim_{x \to 0} \frac{\sec^2 x - \cos x}{3x^2} \]

2. Evaluating this limit again leads to \( \frac{0}{0} \), so we apply L'Hôpital's Rule again:
- The derivative of \( \sec^2 x \) is \( 2 \sec^2 x \tan x \).
- The derivative of \( \cos x \) is \( -\sin x \).
- The derivative of \( 3x^2 \) is \( 6x \).

Thus:
\[ \text{RHL} = \lim_{x \to 0} \frac{2 \sec^2 x \tan x + \sin x}{6x} \]

3. Evaluating this limit again leads to \( \frac{0}{0} \), so we apply L'Hôpital's Rule one more time:
- The derivative of \( 2 \sec^2 x \tan x \) and \( \sin x \) will yield a more complex expression, but we can simplify using Taylor series or known limits.

Using Taylor series expansion:
\[ \tan x \approx x + \frac{x^3}{3} + O(x^5) \quad \text{and} \quad \sin x \approx x - \frac{x^3}{6} + O(x^5) \]
Thus:
\[ \tan x - \sin x \approx \left(x + \frac{x^3}{3}\right) - \left(x - \frac{x^3}{6}\right) = \frac{x^3}{3} + \frac{x^3}{6} = \frac{1}{2}x^3 \]
So:
\[ \text{RHL} = \lim_{x \to 0} \frac{\frac{1}{2}x^3}{x^3} = \frac{1}{2} \]

### Step 4: Set LHL equal to RHL
Now we set the LHL equal to the RHL:
\[ a = \frac{1}{2} \]

### Final Answer
Thus, the value of \( a \) for which the function \( f \) is continuous at \( x = 0 \) is:
\[ \boxed{\frac{1}{2}} \]