`2"tan"(tan^(-1)(x)+tan^(-1)(x^3)),w h e r e x in R-{-1,1},` is equal to `(2x)/(1-x^2)` `t(2tan^(-1)x)` `tan(cot^(-1)(-x)-cot^(-1)(x))` `"tan"(2cot^(-1)x)`

Options

A. `(2x)/(1 -x^(2))`
B. `tan(2 tan^(-1) x)`
C. `tan (cot^(-1) (-x) - cot^(-1) x))`
D. `tan (2 cot^(-1) x)`

Explanation

To solve the problem \( 2 \tan(\tan^{-1}(x) + \tan^{-1}(x^3)) \), where \( x \in \mathbb{R} \setminus \{-1, 1\} \), we can follow these steps:

### Step 1: Define Variables
Let \( \alpha = \tan^{-1}(x) \) and \( \beta = \tan^{-1}(x^3) \). Then, we have:
- \( \tan(\alpha) = x \)
- \( \tan(\beta) = x^3 \)

**Hint:** Start by expressing the inverse tangent functions in terms of variables to simplify the expression.

### Step 2: Use the Sum Formula for Tangent
We need to find \( \tan(\alpha + \beta) \). The formula for the tangent of a sum is:
\[ \tan(\alpha + \beta) = \frac{\tan(\alpha) + \tan(\beta)}{1 - \tan(\alpha) \tan(\beta)} \]
Substituting the values of \( \tan(\alpha) \) and \( \tan(\beta) \):
\[ \tan(\alpha + \beta) = \frac{x + x^3}{1 - x \cdot x^3} \]

**Hint:** Remember to apply the tangent addition formula correctly.

### Step 3: Simplify the Expression
Now simplify the expression:
\[ \tan(\alpha + \beta) = \frac{x + x^3}{1 - x^4} \]
Thus, we have:
\[ 2 \tan(\alpha + \beta) = 2 \cdot \frac{x + x^3}{1 - x^4} \]
This simplifies to:
\[ \frac{2(x + x^3)}{1 - x^4} \]

**Hint:** Factor out common terms in the numerator to simplify further.

### Step 4: Factor the Numerator
Notice that:
\[ x + x^3 = x(1 + x^2) \]
So we can rewrite:
\[ 2 \tan(\alpha + \beta) = \frac{2x(1 + x^2)}{1 - x^4} \]

**Hint:** Look for ways to factor the denominator as well.

### Step 5: Factor the Denominator
The denominator can be factored as:
\[ 1 - x^4 = (1 - x^2)(1 + x^2) \]
Thus, we have:
\[ 2 \tan(\alpha + \beta) = \frac{2x(1 + x^2)}{(1 - x^2)(1 + x^2)} \]

**Hint:** Cancel out the common terms in the numerator and denominator.

### Step 6: Cancel Common Terms
The \( (1 + x^2) \) terms cancel out:
\[ 2 \tan(\alpha + \beta) = \frac{2x}{1 - x^2} \]

### Conclusion
Thus, we find that:
\[ 2 \tan(\tan^{-1}(x) + \tan^{-1}(x^3)) = \frac{2x}{1 - x^2} \]

The answer is:
\[ \boxed{\frac{2x}{1 - x^2}} \]