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 \( a \) so 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 compute

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

### Step 2: Rewrite \( \tan x \) and \( \cot x \)

Using the definitions of tangent and cotangent, we have:

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

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 substituting this into the limit gives:

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

### Step 4: Factor the numerator

We can factor the numerator using the identity \( \sin^2 x - \cos^2 x = -\cos(2x) \):

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

### Step 5: Apply 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: \( \frac{d}{dx}(-\cos(2x)) = 2\sin(2x) \).
2. Differentiate the denominator: Using the product rule on \( (x - \frac{\pi}{4}) \sin x \cos x \).

### Step 6: Evaluate the limit

After applying L'Hôpital's Rule, we can evaluate the limit again. However, to simplify the calculations, we can also use the Taylor series expansion around \( x = \frac{\pi}{4} \) for \( \sin x \) and \( \cos x \) to find that:

\[ \sin x \approx \frac{1}{\sqrt{2}}, \quad \cos x \approx \frac{1}{\sqrt{2}} \quad \text{at } x = \frac{\pi}{4}. \]

Thus, substituting back into the limit:

\[ \lim_{x \to \frac{\pi}{4}} \frac{-\cos(2(\frac{\pi}{4}))}{(x - \frac{\pi}{4}) \cdot \frac{1}{2}} = \lim_{x \to \frac{\pi}{4}} \frac{-0}{(x - \frac{\pi}{4}) \cdot \frac{1}{2}} = 0. \]

### Step 7: Set the limit equal to \( a \)

To ensure continuity at \( x = \frac{\pi}{4} \):

\[ a = \lim_{x \to \frac{\pi}{4}} f(x) = 0. \]

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

\[ \boxed{0}. \]