`tan^(- 1)x+tan^(- 1)\ (2x)/(1-x^2)=pi+tan^(- 1)\ (3x-x^3)/(1-3x^2),(x >0)` is true if

Options

A. `x lt (1)/(sqrt3)`
B. `(1)/(sqrt3) lt x lt 1`
C. `x gt (1)/(sqrt3)`
D. `(1)/(sqrt3) lt x lt sqrt3`

Answer

To solve the equation

\[ \tan^{-1}(x) + \tan^{-1}\left(\frac{2x}{1-x^2}\right) = \pi + \tan^{-1}\left(\frac{3x-x^3}{1-3x^2}\right), \quad (x > 0) \]

we can use the properties of the inverse tangent function.

### Step 1: Use the formula for the tangent of a sum
We know that:

\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a + b}{1 - ab}\right) \]

when \(ab < 1\).

In our case, we can apply this to the left-hand side:

Let \(a = x\) and \(b = \frac{2x}{1-x^2}\).

Then,

\[ \tan^{-1}(x) + \tan^{-1}\left(\frac{2x}{1-x^2}\right) = \tan^{-1}\left(\frac{x + \frac{2x}{1-x^2}}{1 - x \cdot \frac{2x}{1-x^2}}\right) \]

### Step 2: Simplify the left-hand side
Now, we simplify the expression:

The numerator becomes:

\[ x + \frac{2x}{1-x^2} = \frac{x(1-x^2) + 2x}{1-x^2} = \frac{x(1 - x^2 + 2)}{1-x^2} = \frac{x(3 - x^2)}{1-x^2} \]

The denominator becomes:

\[ 1 - x \cdot \frac{2x}{1-x^2} = \frac{(1-x^2) - 2x^2}{1-x^2} = \frac{1 - 3x^2}{1-x^2} \]

Thus, we have:

\[ \tan^{-1}\left(\frac{x(3 - x^2)}{1 - 3x^2}\right) \]

So, the left-hand side simplifies to:

\[ \tan^{-1}\left(\frac{x(3 - x^2)}{1 - 3x^2}\right) \]

### Step 3: Set the left-hand side equal to the right-hand side
Now we equate this to the right-hand side:

\[ \tan^{-1}\left(\frac{x(3 - x^2)}{1 - 3x^2}\right) = \pi + \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right) \]

### Step 4: Use the properties of the inverse tangent
Using the property \(\tan^{-1}(a) + \pi = \tan^{-1}(a) + \tan^{-1}(-1)\) when \(a < 0\), we can rewrite the right-hand side:

\[ \tan^{-1}\left(\frac{x(3 - x^2)}{1 - 3x^2}\right) = \tan^{-1}\left(\frac{3x - x^3}{1 - 3x^2}\right) + \pi \]

This implies:

\[ \frac{x(3 - x^2)}{1 - 3x^2} = -\frac{3x - x^3}{1 - 3x^2} \]

### Step 5: Solve the equation
Cross-multiplying gives:

\[ x(3 - x^2) = -(3x - x^3) \]

This simplifies to:

\[ x(3 - x^2) + 3x - x^3 = 0 \]

Combining terms:

\[ 3x - x^3 + 3x - x^3 = 0 \implies 6x - 2x^3 = 0 \]

Factoring out \(2x\):

\[ 2x(3 - x^2) = 0 \]

Thus, \(x = 0\) or \(3 - x^2 = 0\) which gives \(x^2 = 3\) or \(x = \sqrt{3}\).

### Step 6: Determine the valid range for \(x\)
Since \(x > 0\), we have \(x = \sqrt{3}\).

Now we need to check the conditions under which the original equation holds. From the properties of the tangent function, we find that the equation is valid when \(x\) lies between \(\frac{1}{\sqrt{3}}\) and \(1\).

### Conclusion
Thus, the solution to the equation is:

\[ x \in \left(\frac{1}{\sqrt{3}}, 1\right) \]