91Ó°ÊÓ

We are given the function $$y = \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right).$$ Our first step is to break down the function into simpler forms, allowing us to better analyze each part of the function. In this case, we break down the expression inside the inverse sine function: $$\frac{1-x^2}{1+x^2}.$$

The inverse sine function, denoted by $$\sin^{-1}$$, is the inverse of the sine function. When we apply the inverse sine function to a value, we are finding the angle whose sine value is that given value. In our function, we have $$y = \sin^{-1}\left(\frac{1-x^2}{1+x^2}\right).$$ This means we are finding the angle y whose sine value is equal to $$\frac{1-x^2}{1+x^2}$$. Also, keep in mind that the domain of the inverse sine function is $$[-1, 1]$$, and its range is $$\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$$. This implies that the given fraction must lie within $$[-1, 1]$$ for it to be possible to calculate the inverse sine for it.

As mentioned in the previous step, the domain of the inverse sine function is $$[-1, 1]$$. Therefore, we need to check if $$\frac{1-x^2}{1+x^2}$$ falls within this domain for any possible value of x. Let's investigate the numerator and the denominator of the fraction separately. For the numerator, we have $$1 - x^2$$. This can be any value between -∞ to 1 depending on x. Notably, it can be negative or positive, so we cannot determine if it falls within the required domain only based on the numerator. For the denominator, we have $$1 + x^2$$, which will always be a positive value due to the square term ($$x^2$$). However, we don't know yet if the denominator can make the overall fraction fall in the required domain. Now consider the fraction as a whole: $$\frac{1 - x^2}{1 + x^2}$$. This fraction can be positive or negative for different values of x, because both the numerator and the denominator include x and its square. However, for the inverse sine function, it is crucial that this fraction falls within the domain [-1, 1]. As the denominator is always positive, the fraction will have the same sign as the numerator. Thus, the fraction will be in the domain [-1, 1] (acceptable for the inverse sine function) whenever the numerator, $$1-x^2$$, falls between -1 and 1. As the original function is already given, there is no need to solve for any particular values of x that make the fraction fall within the acceptable domain. The main takeaway is that this fraction and, therefore, the original function are indeed valid for the inverse sine function.