If `sin ^(-1)((1 - x ^(2))/(1 + x ^(2)))+ cos ^(-1) ((2x )/(1 +x ^(2)))= pi/3,` then the vlaue of x is equal to

Options

A. `1 //sqrt2`
B. `1 //sqrt3`
C. `sqrt2`
D. `sqrt3`

Answer

To solve the equation

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

we can follow these steps:

### Step 1: Use the identity for sine and cosine
We know that if we let \( x = \tan\left(\frac{y}{2}\right) \), then we can express sine and cosine in terms of \( y \):
- \(\sin y = \frac{2\tan\left(\frac{y}{2}\right)}{1 + \tan^2\left(\frac{y}{2}\right)} = \frac{2x}{1 + x^2}\)
- \(\cos y = \frac{1 - \tan^2\left(\frac{y}{2}\right)}{1 + \tan^2\left(\frac{y}{2}\right)} = \frac{1 - x^2}{1 + x^2}\)

### Step 2: Rewrite the equation
Substituting these into the original equation gives us:

\[ \sin^{-1}\left(\cos y\right) + \cos^{-1}\left(\sin y\right) = \frac{\pi}{3}. \]

### Step 3: Simplify using inverse identities
Using the identity \(\sin^{-1}(a) + \cos^{-1}(a) = \frac{\pi}{2}\), we can rewrite the equation as:

\[ \frac{\pi}{2} - y + \frac{\pi}{2} - y = \frac{\pi}{3}. \]

### Step 4: Combine terms
This simplifies to:

\[ \pi - 2y = \frac{\pi}{3}. \]

### Step 5: Solve for \(y\)
Rearranging gives:

\[ 2y = \pi - \frac{\pi}{3} = \frac{3\pi}{3} - \frac{\pi}{3} = \frac{2\pi}{3}. \]

Dividing by 2, we find:

\[ y = \frac{\pi}{3}. \]

### Step 6: Find \(x\)
Recall that \(x = \tan\left(\frac{y}{2}\right)\):

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

### Step 7: Calculate \(\tan\left(\frac{\pi}{6}\right)\)
We know that:

\[ \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}}. \]

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

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

### Final Answer
The value of \(x\) is

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

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