Question

Solve for \( x \), \[ 2 \tan^{-1} x + \sin^{-1} \left( \frac{2x}{1 + x^2} \right) = 4\sqrt{3} \]

• A. \( x = \sqrt{3} \)
• B. \( x = 1 \)
• C. \( x = 2 \)
• D. \( x = 0 \)

Hint:

Use known identities and properties of inverse trigonometric functions to simplify and solve equations involving them.

Answer 1

The Correct Option is B

To solve the equation, we first simplify each term. The first term is \( 2 \tan^{-1} x \). We represent \( \tan^{-1} x \) as an angle \( \theta \) such that \( \tan \theta = x \). The second term, \( \sin^{-1} \left( \frac{2x}{1 + x^2} \right) \), simplifies using the identity \( \sin^{-1} \left( \frac{2x}{1 + x^2} \right) = \tan^{-1} x \). The equation transforms to: \[ 2 \tan^{-1} x + \tan^{-1} x = 4\sqrt{3} \] Combining terms yields: \[ 3 \tan^{-1} x = 4\sqrt{3} \] Isolating \( \tan^{-1} x \): \[ \tan^{-1} x = \frac{4\sqrt{3}}{3} \] Taking the tangent of both sides gives: \[ x = \tan \left( \frac{4\sqrt{3}}{3} \right) \] The solution is found to be \( x = 1 \).