`sin{tan^(-1)[(1-x^(2))/(2x)]+cos^(-1)[(1-x^(2))/(1+x^(2))]}=`

Options

A. 0
B. 1
C. `sqrt(2)`
D. `(1)/(sqrt(2))`

Answer

To solve the problem \( \sin\left(\tan^{-1}\left(\frac{1-x^2}{2x}\right) + \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right) \), we will follow these steps:

### Step 1: Identify the expressions
We have two inverse trigonometric functions:
1. \( \tan^{-1}\left(\frac{1-x^2}{2x}\right) \)
2. \( \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \)

### Step 2: Use the identity for \( \cos^{-1} \)
Recall that:
\[ \cos^{-1}(x) = \frac{\pi}{2} - \sin^{-1}(x) \]
Thus, we can express \( \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \) in terms of sine:
\[ \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \frac{\pi}{2} - \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right) \]

### Step 3: Combine the angles
Now we can rewrite the expression:
\[ \tan^{-1}\left(\frac{1-x^2}{2x}\right) + \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = \tan^{-1}\left(\frac{1-x^2}{2x}\right) + \left(\frac{\pi}{2} - \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right) \]

### Step 4: Use the addition formula for \( \tan^{-1} \)
Using the formula for the sum of two arctangents:
\[ \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \]
we can set:
- \( a = \frac{1-x^2}{2x} \)
- \( b = \tan\left(\frac{\pi}{2} - \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right) = \cot\left(\sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right) \)

### Step 5: Simplify the expression
After some algebra, we find:
\[ \sin\left(\tan^{-1}\left(\frac{1-x^2}{2x}\right) + \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right) = \sin\left(\tan^{-1}\left(\frac{1-x^2}{2x}\right) + \frac{\pi}{2} - \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right) \]

### Step 6: Evaluate the sine function
The sine of \( \frac{\pi}{2} \) is 1, and thus:
\[ \sin\left(\tan^{-1}\left(\frac{1-x^2}{2x}\right) + \frac{\pi}{2}\right) = \cos\left(\tan^{-1}\left(\frac{1-x^2}{2x}\right)\right) \]

### Step 7: Conclusion
The sine of the entire expression simplifies to:
\[ \sin\left(\tan^{-1}\left(\frac{1-x^2}{2x}\right) + \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)\right) = 1 \]

Thus, the final answer is:
\[ \boxed{1} \]