Let `f(x)=sin^(-1)((2x)/(1+x^(2)))` and `g(x)=cos^(-1)((x^(2)-1)/(x^(2)+1))`. Then tha value of f(10)-g(100) is equal to

Options

A. `pi-2(tan^(-1)(10)+tan^(-1)(100))`
B. 0
C. `2(tan^(-1)(100)-tan^(-1)(10))`
D. `2(tan^(-1)(10)-tan^(-1)(100))`

Explanation

To solve the problem, we need to evaluate \( f(10) - g(100) \) where:

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

### Step 1: Evaluate \( f(10) \)

Using the known formula for \( \sin^{-1} \left( \frac{2x}{1+x^2} \right) \):

\[ f(x) = \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \tan^{-1}(x) \quad \text{for } x \geq 0 \]

Thus, we have:

\[ f(10) = \tan^{-1}(10) \]

### Step 2: Evaluate \( g(100) \)

Using the known formula for \( \cos^{-1} \left( \frac{x^2-1}{x^2+1} \right) \):

\[ g(x) = \cos^{-1}\left(\frac{x^2-1}{x^2+1}\right) = \frac{\pi}{2} - \tan^{-1}(x) \quad \text{for } x \geq 0 \]

Thus, we have:

\[ g(100) = \frac{\pi}{2} - \tan^{-1}(100) \]

### Step 3: Calculate \( f(10) - g(100) \)

Now we can substitute our results into the expression \( f(10) - g(100) \):

\[ f(10) - g(100) = \tan^{-1}(10) - \left(\frac{\pi}{2} - \tan^{-1}(100)\right) \]

This simplifies to:

\[ f(10) - g(100) = \tan^{-1}(10) - \frac{\pi}{2} + \tan^{-1}(100) \]

### Step 4: Combine the terms

Using the identity \( \tan^{-1}(a) + \tan^{-1}(b) = \tan^{-1}\left(\frac{a+b}{1-ab}\right) \) when \( ab < 1 \):

Here, \( 10 \cdot 100 = 1000 > 1 \), so we can write:

\[ \tan^{-1}(10) + \tan^{-1}(100) = \tan^{-1}\left(\frac{10 + 100}{1 - 10 \cdot 100}\right) + \pi \]

Thus, we have:

\[ f(10) - g(100) = \tan^{-1}(10) + \tan^{-1}(100) - \frac{\pi}{2} = \tan^{-1}\left(\frac{110}{-9999}\right) + \pi \]

### Final Result

Since \( \tan^{-1}(x) + \tan^{-1}(y) - \frac{\pi}{2} \) results in \( \tan^{-1}\left(\frac{110}{-9999}\right) + \pi \), we can conclude:

\[ f(10) - g(100) = 2 \]

### Conclusion

Thus, the value of \( f(10) - g(100) \) is \( 2 \).

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