In the provided image, the expression appears to involve inverse trigonometric functions, specifically the sine and cosine inverse functions. Here's a breakdown of what I can see:

The expression includes:

1. $\sin^{-1} \left( \frac{2x}{1 + x^2} \right)$
2. $\cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)$

These expressions often appear in problems that involve the sum or difference of inverse trigonometric functions. The form of $\frac{2x}{1 + x^2}$ and $\frac{1 - x^2}{1 + x^2}$ are typical transformations for expressing $\tan^{-1}$ or combinations thereof.

Possible Approach to Solve or Simplify

The expressions $\sin^{-1} \left( \frac{2x}{1 + x^2} \right)$ and $\cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right)$ can often be simplified because they represent forms related to angle addition identities. Specifically:

• $\frac{2x}{1 + x^2} = \sin(2 \tan^{-1}(x))$
• $\frac{1 - x^2}{1 + x^2} = \cos(2 \tan^{-1}(x))$

Thus: $\sin^{-1} \left( \frac{2x}{1 + x^2} \right) + \cos^{-1} \left( \frac{1 - x^2}{1 + x^2} \right) = \frac{\pi}{2}$

Explanation

Since $\sin^{-1} (x) + \cos^{-1} (x) = \frac{\pi}{2}$ for any argument $x$ within the domain, this property simplifies the entire expression to $\frac{\pi}{2}$.

Would you like further details on this solution, or do you have additional questions about specific parts of the problem?

Related Questions

1. How do you derive the identities for $\frac{2x}{1 + x^2}$ and $\frac{1 - x^2}{1 + x^2}$ in terms of trigonometric functions?
2. What is the domain of the expression $\sin^{-1} \left( \frac{2x}{1 + x^2} \right)$?
3. Can we use a similar approach to solve expressions involving $\tan^{-1}$?
4. How does the identity $\sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2}$ work for complex numbers?
5. What other trigonometric identities are helpful in simplifying inverse trigonometric expressions?

Tip: When working with inverse trigonometric functions, remember their range constraints, as they define the limits of possible simplifications.