Problem
If \( x > 1 \), then express:
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) \] in terms of \( \tan^{-1}x \).
Solution
Let
\[ \tan^{-1}x = \theta \]
Then,
\[ x = \tan \theta \]
Using identity:
\[ \sin 2\theta = \frac{2\tan\theta}{1+\tan^2\theta} \]
Substitute \( x = \tan \theta \):
\[ \frac{2x}{1+x^2} = \sin 2\theta \]
So,
\[ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \sin^{-1}(\sin 2\theta) \]
Now, since \( x > 1 \Rightarrow \theta = \tan^{-1}x > \frac{\pi}{4} \),
\[ 2\theta > \frac{\pi}{2} \]
We know the principal value rule:
\[ \sin^{-1}(\sin y) = \pi – y \quad \text{for } \frac{\pi}{2} < y < \frac{3\pi}{2} \]
Thus,
\[ \sin^{-1}(\sin 2\theta) = \pi – 2\theta \]
Substitute back \( \theta = \tan^{-1}x \):
Final Answer:
\[ \boxed{ \sin^{-1}\left(\frac{2x}{1+x^2}\right) = \pi – 2\tan^{-1}x } \]
Key Concept
Normally, \( \sin^{-1}\left(\frac{2x}{1+x^2}\right) = 2\tan^{-1}x \), but this holds only when the angle lies in principal range. For \( x > 1 \), adjustment is required due to range of sine inverse. :contentReference[oaicite:0]{index=0}