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

Step 1: Understanding the Concept:
Standard inverse trigonometric substitution identities can convert all terms into \(\tan^{-1} x\).
Step 2: Key Formula or Approach:
For \(|x| \le 1\):
1. \(\sin^{-1}(\frac{2x}{1+x^2}) = 2\tan^{-1}x\).
2. \(\cos^{-1}(\frac{1-x^2}{1+x^2}) = 2\tan^{-1}x\).
3. \(\tan^{-1}(\frac{2x}{1-x^2}) = 2\tan^{-1}x\).
Step 3: Detailed Explanation:
Substitute the identities into the given equation:
\[ 3(2\tan^{-1}x) - 4(2\tan^{-1}x) + 2(2\tan^{-1}x) = \frac{\pi}{3} \] \[ 6\tan^{-1}x - 8\tan^{-1}x + 4\tan^{-1}x = \frac{\pi}{3} \] \[ 2\tan^{-1}x = \frac{\pi}{3} \] \[ \tan^{-1}x = \frac{\pi}{6} \] Taking \(\tan\) on both sides:
\[ x = \tan(\frac{\pi}{6}) = \frac{1}{\sqrt{3}} \] Step 4: Final Answer:
The value of \(x\) is \(\frac{1}{\sqrt{3}}\).