If `f(x)=((1-tanx)/(1+sinx))^(cosec x)` is to be made continuous at `x=0,` then `f(0)` must be equal to

Options

A. `e^(2)`
B. `e`
C. `(I)/(e)`
D. `(I)/(e^(2))`

Explanation

To determine the value of \( f(0) \) for the function \( f(x) = \left( \frac{1 - \tan x}{1 + \sin x} \right)^{\csc x} \) to be continuous at \( x = 0 \), we need to ensure that \( f(0) \) equals the limit of \( f(x) \) as \( x \) approaches 0.

### Step 1: Evaluate \( f(0) \)
First, we need to find \( f(0) \):
\[ f(0) = \left( \frac{1 - \tan(0)}{1 + \sin(0)} \right)^{\csc(0)} \]
Since \( \tan(0) = 0 \) and \( \sin(0) = 0 \), we have:
\[ f(0) = \left( \frac{1 - 0}{1 + 0} \right)^{\csc(0)} = 1^{\csc(0)} \]
However, \( \csc(0) \) is undefined because \( \csc x = \frac{1}{\sin x} \) and \( \sin(0) = 0 \). Thus, \( f(0) \) is not directly computable.

### Step 2: Find the limit of \( f(x) \) as \( x \to 0 \)
Next, we need to compute the limit:
\[ \lim_{x \to 0} f(x) = \lim_{x \to 0} \left( \frac{1 - \tan x}{1 + \sin x} \right)^{\csc x} \]
This is an indeterminate form of type \( 1^\infty \). To resolve this, we can use the logarithmic limit approach.

### Step 3: Take the natural logarithm
Let:
\[ L = \lim_{x \to 0} f(x) = \lim_{x \to 0} \left( \frac{1 - \tan x}{1 + \sin x} \right)^{\csc x} \]
Taking the natural logarithm:
\[ \ln L = \lim_{x \to 0} \csc x \cdot \ln \left( \frac{1 - \tan x}{1 + \sin x} \right) \]

### Step 4: Simplify the logarithmic expression
Using the small angle approximations \( \tan x \approx x \) and \( \sin x \approx x \) as \( x \to 0 \):
\[ \frac{1 - \tan x}{1 + \sin x} \approx \frac{1 - x}{1 + x} = \frac{1 - x}{1 + x} \]
Now, we can compute:
\[ \ln \left( \frac{1 - x}{1 + x} \right) = \ln(1 - x) - \ln(1 + x) \]
Using Taylor expansion:
\[ \ln(1 - x) \approx -x \quad \text{and} \quad \ln(1 + x) \approx x \]
Thus:
\[ \ln \left( \frac{1 - \tan x}{1 + \sin x} \right) \approx -x - x = -2x \]

### Step 5: Substitute back into the limit
Now substituting back into the limit:
\[ \ln L = \lim_{x \to 0} \csc x \cdot (-2x) = \lim_{x \to 0} -2 \cdot \frac{x}{\sin x} \]
As \( x \to 0 \), \( \frac{x}{\sin x} \to 1 \):
\[ \ln L = -2 \]
Thus:
\[ L = e^{-2} = \frac{1}{e^2} \]

### Conclusion: Value of \( f(0) \)
For \( f(x) \) to be continuous at \( x = 0 \), we must have:
\[ f(0) = \lim_{x \to 0} f(x) = \frac{1}{e^2} \]

### Final Answer
Therefore, \( f(0) \) must be equal to \( \frac{1}{e^2} \).

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