Question

If \[ 3\sin^{-1}\left(\frac{2x}{1+x^2}\right) - 4\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) + 2\tan^{-1}\left(\frac{2x}{1-x^2}\right) = \frac{\pi}{3}, \] then the value of \[ x = \, ? \] 

• A. \(\sqrt{3}\)
• B. \(1\)
• C. \(\frac{1}{\sqrt{3}}\)
• D. \(\frac{1}{\sqrt{2}}\)

Hint:

Expressions involving \[ \frac{2x}{1+x^2},\quad \frac{1-x^2}{1+x^2},\quad \frac{2x}{1-x^2} \] almost always suggest the substitution \(x=\tan\theta\).

Answer 1

The Correct Option is C

Concept:
Use the standard substitution: \[ x=\tan\theta \] Then the well-known identities are: \[ \sin^{-1}\left(\frac{2x}{1+x^2}\right)=2\theta \] \[ \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)=2\theta \] \[ \tan^{-1}\left(\frac{2x}{1-x^2}\right)=2\theta \] for appropriate principal values. ip

Step 1: Set \(x=\tan\theta\).
Then the equation becomes: \[ 3(2\theta)-4(2\theta)+2(2\theta)=\frac{\pi}{3} \] ip

Step 2: Simplify the equation.
\[ 6\theta - 8\theta + 4\theta = \frac{\pi}{3} \] \[ 2\theta = \frac{\pi}{3} \] \[ \theta = \frac{\pi}{6} \] ip

Step 3: Find \(x\).
\[ x=\tan\theta=\tan\frac{\pi}{6}=\frac{1}{\sqrt{3}} \] ip Hence, the correct answer is:
\[ \boxed{(C)\ \frac{1}{\sqrt{3}}} \]