`sin [ tan^(-1). (1 - x^(2))/(2x) + cos^(-1) . (1-x^(2))/(1 + x^(2))]` is

Options

A. 1
B. 0
C. `-1`
D. None of these

Answer

To solve the expression \( \sin \left[ \tan^{-1} \left( \frac{1 - x^2}{2x} \right) + \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \right] \), we will use some trigonometric identities and properties of inverse trigonometric functions.

### Step-by-Step Solution:

1. **Identify the first term**:
We start with the term \( \tan^{-1} \left( \frac{1 - x^2}{2x} \right) \). We can use the identity:
\[ \tan^{-1} \left( \frac{2x}{1 - x^2} \right) = 2 \tan^{-1}(x) \]
Therefore, we can rewrite:
\[ \tan^{-1} \left( \frac{1 - x^2}{2x} \right) = \tan^{-1} \left( \frac{1}{x} \right) = \cot^{-1}(x) \]

2. **Identify the second term**:
Next, we consider the term \( \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \). We can use the identity:
\[ \cos^{-1}(y) = \frac{\pi}{2} - \sin^{-1}(y) \]
However, we will also recognize that:
\[ \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) = 2 \tan^{-1}(x) \]

3. **Combine the terms**:
Now we have:
\[ \tan^{-1} \left( \frac{1 - x^2}{2x} \right) + \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) = \cot^{-1}(x) + 2 \tan^{-1}(x) \]

4. **Use the identity for sum of angles**:
We know that:
\[ \cot^{-1}(x) + \tan^{-1}(x) = \frac{\pi}{2} \]
Thus:
\[ \cot^{-1}(x) + 2 \tan^{-1}(x) = \frac{\pi}{2} + \tan^{-1}(x) \]

5. **Evaluate the sine**:
Now we need to find:
\[ \sin \left( \frac{\pi}{2} + \tan^{-1}(x) \right) \]
Using the sine addition formula:
\[ \sin \left( \frac{\pi}{2} + \theta \right) = \cos(\theta) \]
Therefore:
\[ \sin \left( \frac{\pi}{2} + \tan^{-1}(x) \right) = \cos(\tan^{-1}(x)) \]

6. **Find the cosine**:
We know that:
\[ \cos(\tan^{-1}(x)) = \frac{1}{\sqrt{1 + x^2}} \]

### Final Result:
Thus, the final result is:
\[ \sin \left[ \tan^{-1} \left( \frac{1 - x^2}{2x} \right) + \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) \right] = \frac{1}{\sqrt{1 + x^2}} \]