The range of the function ,
`f (x) = tan^(-1) ((1+x)/(1 - x)) - tan^(-1) x ` is

Options

A. `{pi//4}`
B. `{-(pi//4),3pi//4}`
C. `{pi//4, - (3pi//4)}`
D. `{3 pi//4}`

Answer

To find the range of the function \( f(x) = \tan^{-1} \left( \frac{1+x}{1-x} \right) - \tan^{-1}(x) \), we can follow these steps:

### Step 1: Use the property of inverse tangent
We can use the identity for the difference of two inverse tangent functions:
\[ \tan^{-1}(a) - \tan^{-1}(b) = \tan^{-1}\left(\frac{a - b}{1 + ab}\right) \]
In our case, let \( a = \frac{1+x}{1-x} \) and \( b = x \). Thus, we can rewrite \( f(x) \) as:
\[ f(x) = \tan^{-1}\left(\frac{\frac{1+x}{1-x} - x}{1 + \left(\frac{1+x}{1-x}\right)x}\right) \]

### Step 2: Simplify the expression
Now, we need to simplify the expression inside the tangent inverse:
1. Calculate \( \frac{1+x}{1-x} - x \):
\[ \frac{1+x}{1-x} - x = \frac{1+x - x(1-x)}{1-x} = \frac{1+x - x + x^2}{1-x} = \frac{1 + x^2}{1-x} \]

2. Calculate \( 1 + \left(\frac{1+x}{1-x}\right)x \):
\[ 1 + \left(\frac{1+x}{1-x}\right)x = 1 + \frac{x(1+x)}{1-x} = \frac{(1-x) + x + x^2}{1-x} = \frac{1 + x^2}{1-x} \]

### Step 3: Substitute back into the function
Now, substituting these results back into \( f(x) \):
\[ f(x) = \tan^{-1}\left(\frac{\frac{1+x^2}{1-x}}{\frac{1+x^2}{1-x}}\right) = \tan^{-1}(1) = \frac{\pi}{4} \]

### Step 4: Determine the range
Since the function simplifies to a constant value \( \frac{\pi}{4} \), the range of \( f(x) \) is simply:
\[ \text{Range of } f(x) = \left\{ \frac{\pi}{4} \right\} \]

### Conclusion
Thus, the range of the function \( f(x) = \tan^{-1} \left( \frac{1+x}{1-x} \right) - \tan^{-1}(x) \) is \( \left\{ \frac{\pi}{4} \right\} \).

---