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\begin{array}{ll} \cos^{-1} \left(\frac{1-x^2}{1+x^2}\right), & \text{if } -1 \leq x \leq 1 \\ \tan^{-1} \left(\frac{2x}{1-x^2}\right), & \text{if } x > 1 \text{ or } x < -1 \end{array} \right.$$ Let’s break this down. ### First Case: $$-1 \leq x \leq 1$$ For $$-1 \leq x \leq 1$$: $$\sin^{-1} \left(\frac{2x}{1+x^2}\right) = \cos^{-1} \left(\frac{1-x^2}{1+x^2}\right)$$ This equation exploits the identity $$\sin^{-1} y + \cos^{-1} y = \frac{\pi}{2}$$. If $$y = \frac{2x}{1+x^2}$$, then: $$\sin^{-1} y = \sin^{-1} \left(\frac{2x}{1+x^2}\right) \quad \text{and} \quad \cos^{-1} y = \cos^{-1} \left(\frac{1-x^2}{1+x^2}\right)$$ These two should add up to $$\frac{\pi}{2}$$, confirming that the equation holds true in this range. ### Second Case: $$x > 1$$ or $$x < -1$$ For $$x > 1$$ or $$x < -1$$: $$\sin^{-1} \left(\frac{2x}{1+x^2}\right) = \tan^{-1} \left(\frac{2x}{1-x^2}\right)$$ In this case, the expression is transformed through the tangent function, applicable when $$x$$ is outside the interval $$[-1, 1]$$. The expression is solved and verified based on the intervals of $$x$$. Would you like more details, or do you have any questions? Here are some related questions: 1. What are the geometric interpretations of the arcsine, arccosine, and arctangent functions? 2. How do the ranges of $$x$$ affect the validity of trigonometric identities? 3. What are the properties of the inverse trigonometric functions? 4. How does the tangent function behave for large values of $$x$$? 5. Can you explore more complex cases involving trigonometric identities? **Tip:** When dealing with inverse trigonometric functions, always consider the domain and range constraints to avoid invalid solutions.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Trigonometry
Inverse trigonometric functions
Formulas
Identity: arcsin(y) + arccos(y) = pi/2
Identity: arctan(y)
Theorems
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Suitable Grade Level
Advanced Mathematics
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