If `3\ sin^(-1)((2x)/(1+x^2))-4\ cos^(-1)((1-x^2)/(1+x^2))+2\ tan^(-1)((2x)/(1-x^2))=pi/3` , then `x` is equal to `1/(sqrt(3))` (b) `-1/(sqrt(3))` (c) `sqrt(3)` (d) `-(sqrt(3))/4`

Options

A. `(1)/(sqrt3)`
B. `-(1)/(sqrt3)`
C. `sqrt3`
D. `-(sqrt3)/(4)`

Explanation

To solve the equation

\[ 3 \sin^{-1}\left(\frac{2x}{1+x^2}\right) - 4 \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) + 2 \tan^{-1}\left(\frac{2x}{1-x^2}\right) = \frac{\pi}{3}, \]

we can use some known identities for inverse trigonometric functions.

### Step 1: Use the identities for inverse trigonometric functions

We know that:

1. \(\sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2 \tan^{-1}(x)\)
2. \(\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = 2 \tan^{-1}(x)\)
3. \(\tan^{-1}\left(\frac{2x}{1-x^2}\right) = 2 \tan^{-1}(x)\)

Using these identities, we can rewrite the equation.

### Step 2: Substitute the identities into the equation

Substituting the identities into the equation gives us:

\[ 3 \cdot 2 \tan^{-1}(x) - 4 \cdot 2 \tan^{-1}(x) + 2 \cdot 2 \tan^{-1}(x) = \frac{\pi}{3}. \]

This simplifies to:

\[ 6 \tan^{-1}(x) - 8 \tan^{-1}(x) + 4 \tan^{-1}(x) = \frac{\pi}{3}. \]

### Step 3: Combine like terms

Combining the terms on the left side, we have:

\[ (6 - 8 + 4) \tan^{-1}(x) = \frac{\pi}{3}. \]

This simplifies to:

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

### Step 4: Solve for \(\tan^{-1}(x)\)

Dividing both sides by 2 gives us:

\[ \tan^{-1}(x) = \frac{\pi}{6}. \]

### Step 5: Find \(x\)

Taking the tangent of both sides, we find:

\[ x = \tan\left(\frac{\pi}{6}\right). \]

Since \(\tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}\), we have:

\[ x = \frac{1}{\sqrt{3}}. \]

### Final Answer

Thus, the value of \(x\) is:

\[ \boxed{\frac{1}{\sqrt{3}}}. \]

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