Question
If \( x > 1 \), evaluate:
\[ 2\tan^{-1}x + \sin^{-1}\left(\frac{2x}{1+x^2}\right) \]
Solution
Use identity:
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x \quad \text{(for } x \le 1\text{)} \]
But for \( x > 1 \), principal value adjustment gives:
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \pi – 2\tan^{-1}x \]
Thus,
\[ 2\tan^{-1}x + \left(\pi – 2\tan^{-1}x\right) = \pi \]
Final Answer:
\[ \boxed{\pi} \]
Key Concept
Adjust identities based on domain and principal value ranges.