Let f(x)=`{{:(,(tanx-cotx)/(x-(pi)/(4)),x ne (pi)/(4)),(,a,x=(pi)/(4)):}`
The value of a so that f(x) is a continous at `x=pi//4` is.

Options

A. 2
B. 4
C. 3
D. 1

Explanation

To find the value of \( a \) such that the function

\[ f(x) = \begin{cases} \frac{\tan x - \cot x}{x - \frac{\pi}{4}} & \text{if } x \neq \frac{\pi}{4} \\ a & \text{if } x = \frac{\pi}{4} \end{cases} \]

is continuous at \( x = \frac{\pi}{4} \), we need to ensure that

\[ \lim_{x \to \frac{\pi}{4}} f(x) = f\left(\frac{\pi}{4}\right) = a. \]

### Step 1: Calculate the limit as \( x \) approaches \( \frac{\pi}{4} \)

We need to find

\[ \lim_{x \to \frac{\pi}{4}} \frac{\tan x - \cot x}{x - \frac{\pi}{4}}. \]

### Step 2: Simplify \( \tan x - \cot x \)

Recall that

\[ \tan x = \frac{\sin x}{\cos x} \quad \text{and} \quad \cot x = \frac{\cos x}{\sin x}. \]

Thus,

\[ \tan x - \cot x = \frac{\sin x}{\cos x} - \frac{\cos x}{\sin x} = \frac{\sin^2 x - \cos^2 x}{\sin x \cos x}. \]

### Step 3: Substitute into the limit

Now we can rewrite the limit:

\[ \lim_{x \to \frac{\pi}{4}} \frac{\frac{\sin^2 x - \cos^2 x}{\sin x \cos x}}{x - \frac{\pi}{4}} = \lim_{x \to \frac{\pi}{4}} \frac{\sin^2 x - \cos^2 x}{(x - \frac{\pi}{4}) \sin x \cos x}. \]

### Step 4: Use L'Hôpital's Rule

Since both the numerator and denominator approach 0 as \( x \to \frac{\pi}{4} \), we can apply L'Hôpital's Rule:

1. Differentiate the numerator:
\[ \frac{d}{dx}(\sin^2 x - \cos^2 x) = 2\sin x \cos x + 2\cos x \sin x = 4\sin x \cos x. \]

2. Differentiate the denominator:
\[ \frac{d}{dx}(x - \frac{\pi}{4}) = 1. \]

Thus, we have:

\[ \lim_{x \to \frac{\pi}{4}} \frac{4\sin x \cos x}{1} = 4\sin\left(\frac{\pi}{4}\right)\cos\left(\frac{\pi}{4}\right). \]

### Step 5: Evaluate the limit

Since \( \sin\left(\frac{\pi}{4}\right) = \cos\left(\frac{\pi}{4}\right) = \frac{1}{\sqrt{2}} \):

\[ \lim_{x \to \frac{\pi}{4}} f(x) = 4 \cdot \frac{1}{\sqrt{2}} \cdot \frac{1}{\sqrt{2}} = 4 \cdot \frac{1}{2} = 2. \]

### Step 6: Set the limit equal to \( a \)

For \( f(x) \) to be continuous at \( x = \frac{\pi}{4} \):

\[ a = \lim_{x \to \frac{\pi}{4}} f(x) = 2. \]

### Final Answer

Thus, the value of \( a \) such that \( f(x) \) is continuous at \( x = \frac{\pi}{4} \) is

\[ \boxed{2}. \]