Statement I Derivative of `sin^(-1)((2x)/(1+x^(2)))w.r.t. cos^(-1)((1-x^(2))/(1+x^(2)))` is 1 for `0ltxlt1.`
`sin^(-1)((2x)/(1+x^(2)))=cos^(-1)((1-x^(2))/(1+x^(2)))` for `-1lexle1`

Options

A. Both statement I and Statement II are correct and Statement II is the correct explanation of Statement I
B. Both Statement I and Statement II are correct but Statement II is not the correct explanation of Statement I
C. Statement I is correct but Statement II is incorrect
D. Statement II is correct but Statement I is incorrect.

Answer

To solve the given problem step by step, we will analyze both statements and derive the necessary results.

### Step 1: Analyze Statement II
We start with Statement II:
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \quad \text{for } -1 \leq x \leq 1. \]

**Hint:** Recall the relationship between sine and cosine functions.

### Step 2: Use the Identity
We know that:
\[ \sin^{-1}(y) + \cos^{-1}(y) = \frac{\pi}{2} \quad \text{for } -1 \leq y \leq 1. \]
Let \( y = \frac{1-x^2}{1+x^2} \). Then:
\[ \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \frac{\pi}{2} - \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right). \]

**Hint:** Use the identity to express one function in terms of the other.

### Step 3: Verify the Equality
To verify the equality, we need to check if:
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) + \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \frac{\pi}{2}. \]

**Hint:** Use the sine addition formula to simplify the expressions.

### Step 4: Simplify the Left Side
Using the sine addition formula:
\[ \sin(A + B) = \sin A \cos B + \cos A \sin B. \]
Let \( A = \sin^{-1}\left(\frac{2x}{1+x^2}\right) \) and \( B = \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right) \).

**Hint:** Find the sine and cosine values for both angles.

### Step 5: Check the Range
For \( -1 \leq x \leq 1 \), both expressions \( \frac{2x}{1+x^2} \) and \( \frac{1-x^2}{1+x^2} \) are valid sine and cosine values.

**Hint:** Ensure that both expressions fall within the range of sine and cosine functions.

### Step 6: Conclusion for Statement II
Since the equality holds true for all \( x \) in the interval \( -1 \leq x \leq 1 \), Statement II is **true**.

### Step 7: Analyze Statement I
Now we analyze Statement I:
\[ \frac{d}{d(\cos^{-1}((1-x^2)/(1+x^2)))}\left(\sin^{-1}\left(\frac{2x}{1+x^2}\right)\right) = 1 \quad \text{for } 0 < x < 1. \]

**Hint:** We will differentiate both sides and check the derivative.

### Step 8: Differentiate the Functions
Let:
\[ f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right), \quad g(x) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right). \]
We need to find \( \frac{df}{dg} \).

**Hint:** Use the chain rule for differentiation.

### Step 9: Apply the Chain Rule
Using the chain rule:
\[ \frac{df}{dg} = \frac{df/dx}{dg/dx}. \]

### Step 10: Compute Derivatives
Calculate \( \frac{df}{dx} \) and \( \frac{dg}{dx} \):
1. For \( f(x) \):
\[ \frac{df}{dx} = \frac{2}{\sqrt{1 - \left(\frac{2x}{1+x^2}\right)^2}} \cdot \frac{(1+x^2)(2) - 2x(2x)}{(1+x^2)^2}. \]
2. For \( g(x) \):
\[ \frac{dg}{dx} = -\frac{2x}{\sqrt{1 - \left(\frac{1-x^2}{1+x^2}\right)^2}} \cdot \frac{(1+x^2)(-2x) - (1-x^2)(2x)}{(1+x^2)^2}. \]

**Hint:** Simplify both derivatives to see if they yield the same result.

### Step 11: Verify the Result
After simplification, if \( \frac{df}{dg} = 1 \), then Statement I is **true**.

### Conclusion
- Statement I is **true**.
- Statement II is **true** but does not explain Statement I.

Thus, the correct option is **C**: Statement I is correct, but Statement II is incorrect.