Question
If
\[ \sin^{-1}\left(\frac{2a}{1+a^2}\right) + \cos^{-1}\left(\frac{1-a^2}{1+a^2}\right) = \tan^{-1}\left(\frac{2x}{1-x^2}\right) \]
where \( a, x \in (0,1) \), find \( x \).
Solution
Let
\[ a = \tan\theta \]
Then,
\[ \frac{2a}{1+a^2} = \sin 2\theta \quad,\quad \frac{1-a^2}{1+a^2} = \cos 2\theta \]
So LHS becomes:
\[ \sin^{-1}(\sin 2\theta) + \cos^{-1}(\cos 2\theta) \]
Since \( a \in (0,1) \Rightarrow \theta \in (0,\frac{\pi}{4}) \Rightarrow 2\theta \in (0,\frac{\pi}{2}) \)
Thus:
\[ \sin^{-1}(\sin 2\theta) = 2\theta \quad,\quad \cos^{-1}(\cos 2\theta) = 2\theta \]
\[ \Rightarrow \text{LHS} = 4\theta \]
Now RHS:
\[ \tan^{-1}\left(\frac{2x}{1-x^2}\right) = 2\tan^{-1}x \]
Thus equation becomes:
\[ 4\theta = 2\tan^{-1}x \Rightarrow 2\theta = \tan^{-1}x \]
\[ x = \tan(2\theta) \]
Using identity:
\[ \tan(2\theta) = \frac{2a}{1-a^2} \]
Final Answer:
\[ \boxed{x = \frac{2a}{1-a^2}} \]
Key Concept
Use substitution \(a = \tan\theta\) and convert into standard double-angle identities.