If `sin^(-1)((2x)/(1+x^(2)))+cos^(-1)((1-x^(2))/(1+x^(2)))=4 tan^(-1) x` then

Options

A. `x in -(-oo,-1)`
B. `x in (1,oo)`
C. `x in [0,1]`
D. `x in [-1,0)`

Answer

To solve the equation

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

we will follow these steps:

### Step 1: Recognize the Inverse Trigonometric Functions
We know that the expressions \(\sin^{-1}\left(\frac{2x}{1+x^2}\right)\) and \(\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\) can be derived from the double angle formulas for tangent. Specifically, we can use the identity:

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

### Step 2: Rewrite the Equation
Using the identity, we can rewrite the left-hand side:

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

Thus, we can rewrite the equation as:

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

### Step 3: Simplify the Equation
Subtract \(2 \tan^{-1}(x)\) from both sides:

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

### Step 4: Use the Identity for Cosine
We know that:

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

Thus, we can rewrite the left-hand side using the sine function:

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

### Step 5: Rearranging the Equation
Rearranging gives us:

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

### Step 6: Use the Sine Function
Taking the sine of both sides, we have:

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

### Step 7: Use the Cosine Double Angle Formula
Using the double angle formula for cosine:

\[ \cos(2\theta) = \frac{1 - \tan^2(\theta)}{1 + \tan^2(\theta)}, \]

we can substitute \(\theta = \tan^{-1}(x)\):

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

### Step 8: Set the Equations Equal
Now we have:

\[ \frac{1-x^2}{1+x^2} = \frac{1-x^2}{1+x^2}. \]

This equation is always true for all \(x\) where the expressions are defined.

### Step 9: Determine the Range of x
Since we have derived that both sides are equal, we need to find the valid range for \(x\). The expressions \(\sin^{-1}\) and \(\cos^{-1}\) are defined for:

- \(\frac{2x}{1+x^2}\) must be in the range \([-1, 1]\).
- \(\frac{1-x^2}{1+x^2}\) must also be in the range \([-1, 1]\).

### Step 10: Find the Intersection of Ranges
The first expression is valid for \(x\) in the interval \([-1, 1]\), and the second expression is valid for \(x\) in the interval \([0, \infty)\). Thus, the intersection of these two ranges is:

\[ x \in [0, 1]. \]

### Final Answer
Therefore, the solution to the equation is:

\[ x \in [0, 1]. \]