Question
Solve:
\[ 3\sin^{-1}\left(\frac{2x}{1+x^2}\right) – 4\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) + 2\tan^{-1}\left(\frac{2x}{1-x^2}\right) = \frac{\pi}{3} \]
Solution
Let
\[ x = \tan\theta \]
Then use standard identities:
\[ \frac{2x}{1+x^2} = \sin 2\theta \]
\[ \frac{1-x^2}{1+x^2} = \cos 2\theta \]
\[ \frac{2x}{1-x^2} = \tan 2\theta \]
So equation becomes:
\[ 3\sin^{-1}(\sin 2\theta) – 4\cos^{-1}(\cos 2\theta) + 2\tan^{-1}(\tan 2\theta) = \frac{\pi}{3} \]
Now use principal values:
- \(\sin^{-1}(\sin 2\theta) = 2\theta\)
- \(\cos^{-1}(\cos 2\theta) = 2\theta\)
- \(\tan^{-1}(\tan 2\theta) = 2\theta\)
(valid when \(2\theta \in [0, \frac{\pi}{2}]\))
Thus:
\[ 3(2\theta) – 4(2\theta) + 2(2\theta) = \frac{\pi}{3} \]
\[ 6\theta – 8\theta + 4\theta = \frac{\pi}{3} \Rightarrow 2\theta = \frac{\pi}{3} \]
\[ \theta = \frac{\pi}{6} \]
Thus,
\[ x = \tan\theta = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \]
Final Answer:
\[ \boxed{\frac{1}{\sqrt{3}}} \]
Key Concept
Use substitution \(x = \tan\theta\) to convert into standard identities.