`2tan^(-1)x = tan^(-1) 2x/(1-x^(2))` is true if

Options

A. `x lt 1`
B. `x gt 1`
C. `x lt 0`
D. `x in R`

Answer

To solve the equation \( 2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \), we will analyze the conditions under which this equality holds true.

### Step 1: Understand the formula for the sum of inverse tangents
We know from trigonometric identities that:
\[ \tan^{-1}a + \tan^{-1}b = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \]
This formula is valid under the condition that \( ab < 1 \).

### Step 2: Apply the formula to our equation
In our case, we can set \( a = x \) and \( b = x \):
\[ \tan^{-1}x + \tan^{-1}x = \tan^{-1}\left(\frac{x+x}{1-xx}\right) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \]
Thus, we can rewrite the left-hand side:
\[ 2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \]

### Step 3: Identify the condition for validity
For the formula to hold, we need:
\[ x \cdot x < 1 \quad \Rightarrow \quad x^2 < 1 \]
This implies:
\[ |x| < 1 \]
Thus, the condition for the equality \( 2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \) to be true is:
\[ x < 1 \quad \text{and} \quad x > -1 \]

### Conclusion
The equation \( 2\tan^{-1}x = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \) is true if \( |x| < 1 \).