If `cos^(-1)((1-x^(2))/(1+x^(2)))+sin^(-1)((2x)/(1+x^(2)))=p` for all `x in[-1,0]`, then p is equal to

Options

A. `(-pi)/(2)`
B. 0
C. `(pi)/(2)`
D. `(2pi)/(3)`

Explanation

To solve the equation

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

for \( x \in [-1, 0] \), we will use the properties of inverse trigonometric functions.

### Step 1: Simplifying the first term

We start with the first term:

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

Using the identity \( \cos^{-1}(y) = \tan^{-1}\left(\frac{\sqrt{1-y^2}}{y}\right) \), we can rewrite it. However, we can also recognize that:

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

Thus,

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

### Step 2: Simplifying the second term

Now, we simplify the second term:

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

Using the identity \( \sin^{-1}(y) = \tan^{-1}\left(\frac{y}{\sqrt{1-y^2}}\right) \), we can also recognize that:

\[ \frac{2x}{1+x^2} = \sin(2\theta) \text{ where } x = \tan(\theta) \]

Thus,

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

### Step 3: Combining the results

Now we can combine the results from Steps 1 and 2:

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

This simplifies to:

\[ 2\tan^{-1}(x) + 2\tan^{-1}(x) = 4\tan^{-1}(x) \]

### Step 4: Finding \( p \)

Since \( x \) ranges from \(-1\) to \(0\), we can evaluate \( \tan^{-1}(x) \):

- When \( x = -1 \), \( \tan^{-1}(-1) = -\frac{\pi}{4} \)
- When \( x = 0 \), \( \tan^{-1}(0) = 0 \)

Thus, \( 4\tan^{-1}(x) \) will range from:

\[ 4\left(-\frac{\pi}{4}\right) = -\pi \quad \text{to} \quad 4(0) = 0 \]

However, since we are looking for a constant value \( p \) that holds for all \( x \) in the interval, we can evaluate the expression at \( x = 0 \):

\[ p = 4\tan^{-1}(0) = 0 \]

### Final Answer

Thus, the value of \( p \) is

\[ \boxed{0} \]