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 will follow these steps:

### Step 1: Use the identity for \(\cos^{-1}(y)\)

We know that

\[ \cos^{-1}(y) = \frac{\pi}{2} - \sin^{-1}(y). \]

Using this identity, we can rewrite the equation:

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

### Step 2: Simplify the equation

Rearranging the equation gives us:

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

This simplifies to:

\[ \sin^{-1}\left(\frac{1 - x^2}{1 + x^2}\right) - \sin^{-1}\left(\frac{2x}{1 + x^2}\right) = -\frac{\pi}{6}. \]

### Step 3: Use the sine difference identity

Using the identity \(\sin^{-1}(a) - \sin^{-1}(b) = \sin^{-1\left(a\sqrt{1-b^2} - b\sqrt{1-a^2}\right)}\), we can rewrite the left-hand side:

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

Thus, we have:

\[ \sin^{-1}\left(\frac{1 - x^2}{1 + x^2}\sqrt{1 - \left(\frac{2x}{1 + x^2}\right)^2} - \frac{2x}{1 + x^2}\sqrt{1 - \left(\frac{1 - x^2}{1 + x^2}\right)^2}\right) = -\frac{\pi}{6}. \]

### Step 4: Solve for \(x\)

To solve for \(x\), we can set:

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

This equation can be quite complex, but we can also approach the problem by substituting values for \(x\) or using trigonometric identities.

### Step 5: Substituting \(x = \tan(y)\)

Let \(x = \tan(y)\). Then, we can express the equation in terms of \(y\):

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

Using the double angle formulas, we can find that:

\[ \sin^{-1}(\cos(2y)) + \cos^{-1}(\sin(2y)) = \frac{\pi}{3}. \]

### Step 6: Final Calculation

From the above, we find that:

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

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

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