If `0< x< 1`,then `tan^(-1)(sqrt(1-x^2)/(1+x))` is equal to

Options

A. `1/2cos^(-1)x`
B. `cos^(-1)sqrt((1+x)/2)`
C. `"Sin"^(-1)sqrt((1-x)/2)`
D. `1/2 sin^(-1)x`

Explanation

To solve the problem, we need to find the value of \( \tan^{-1}\left(\frac{\sqrt{1-x^2}}{1+x}\right) \) given that \( 0 < x < 1 \).

### Step-by-Step Solution:

1. **Set Up the Equation:**
Let \( \theta = \tan^{-1}\left(\frac{\sqrt{1-x^2}}{1+x}\right) \). This implies:
\[ \tan \theta = \frac{\sqrt{1-x^2}}{1+x} \]

2. **Identify the Components:**
Here, we can identify the opposite side (perpendicular) as \( \sqrt{1-x^2} \) and the adjacent side (base) as \( 1+x \).

3. **Calculate the Hypotenuse:**
Using the Pythagorean theorem, the hypotenuse \( h \) can be calculated as:
\[ h = \sqrt{(\sqrt{1-x^2})^2 + (1+x)^2} = \sqrt{(1-x^2) + (1 + 2x + x^2)} = \sqrt{2 + 2x} \]

4. **Simplify the Hypotenuse:**
Factor out the common term:
\[ h = \sqrt{2(1+x)} \]

5. **Find \( \sin \theta \) and \( \cos \theta \):**
- **For \( \sin \theta \):**
\[ \sin \theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{\sqrt{1-x^2}}{\sqrt{2(1+x)}} = \frac{\sqrt{1-x^2}}{\sqrt{2} \sqrt{1+x}} \]

- **For \( \cos \theta \):**
\[ \cos \theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{1+x}{\sqrt{2(1+x)}} = \frac{1+x}{\sqrt{2} \sqrt{1+x}} = \frac{\sqrt{1+x}}{\sqrt{2}} \]

6. **Express \( \theta \) in Terms of Inverse Functions:**
From the sine and cosine values, we can express \( \theta \):
- Using \( \sin \theta \):
\[ \theta = \sin^{-1}\left(\frac{\sqrt{1-x^2}}{\sqrt{2(1+x)}}\right) \]
- Using \( \cos \theta \):
\[ \theta = \cos^{-1}\left(\frac{\sqrt{1+x}}{\sqrt{2}}\right) \]

7. **Final Result:**
Therefore, we conclude that:
\[ \tan^{-1}\left(\frac{\sqrt{1-x^2}}{1+x}\right) = \sin^{-1}\left(\frac{\sqrt{1-x^2}}{\sqrt{2(1+x)}}\right) = \cos^{-1}\left(\frac{\sqrt{1+x}}{\sqrt{2}}\right) \]