`tan^(-1)(cotx)`

Options

A. `f(x)` is periodic with period `pi`
B. f(x) is discontinuous at `x=(pi)/2, (3pi)/2`
C. f(x) is not differential at `x= pi , 99 pi , 100 pi `
D. `f'(x)=-1`, for `2n pi lt x lt (2n+1)pi`

Answer

To solve the problem \( f(x) = \tan^{-1}(\cot x) \), we will follow these steps:

### Step 1: Use the Inverse Trigonometric Identity
We know that:
\[ \tan^{-1}(x) + \cot^{-1}(x) = \frac{\pi}{2} \]
Using this identity, we can rewrite \( f(x) \):
\[ f(x) = \tan^{-1}(\cot x) = \frac{\pi}{2} - \cot^{-1}(\cot x) \]

### Step 2: Simplify \( \cot^{-1}(\cot x) \)
Since \( \cot^{-1}(\cot x) \) is defined as \( x \) when \( x \) is in the range of \( \cot^{-1} \), we have:
\[ \cot^{-1}(\cot x) = x \quad \text{(for } x \text{ in the appropriate range)} \]
Thus, we can express \( f(x) \) as:
\[ f(x) = \frac{\pi}{2} - x \]

### Step 3: Differentiate \( f(x) \)
Now we differentiate \( f(x) \):
\[ f'(x) = \frac{d}{dx}\left(\frac{\pi}{2} - x\right) = 0 - 1 = -1 \]

### Step 4: Analyze the Options
Now we will analyze the options given in the problem:

- **Option A**: \( f(x) \) is periodic.
Since \( f(x) = \frac{\pi}{2} - x \) is a linear function, it is not periodic. **(Incorrect)**

- **Option B**: \( f(x) \) is discontinuous.
The function \( f(x) \) is continuous as it is a polynomial function. **(Incorrect)**

- **Option C**: \( f(x) \) is not differentiable at \( \pi, 99\pi, 100\pi \).
Since \( f'(x) = -1 \) is defined for all \( x \), it is differentiable everywhere. **(Incorrect)**

- **Option D**: The correct option.
Since \( f'(x) = -1 \) is defined everywhere, this is the correct conclusion. **(Correct)**

### Final Answer
The correct option is **D**.

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