Question

For \(|x|\le1\), \(2\tan^{-1}x=\ \ ?\)

• A. \(\tan^{-1}(2x)\)
• B. \(\sin^{-1}\!\left(\dfrac{2x}{1+x^2}\right)\)
• C. \(\cos^{-1}\!\left(\dfrac{2x}{1+x^2}\right)\)
• D. \(\tan^{-1}\!\left(\dfrac{2x}{1+x^2}\right)\)

Hint:

Set $x=\tan\theta$ and convert using double-angle formulas.

Answer 1

The Correct Option is B

Let \(\theta=\tan^{-1}x $\Rightarrow$ x=\tan\theta\) with \(\theta\in[-\pi/4,\pi/4]\). Use $\sin(2\theta)=\dfrac{2\tan\theta}{1+\tan^2\theta}$: \[ \sin(2\theta)=\frac{2x}{1+x^2}. \] Since \(2\theta\in[-\pi/2,\pi/2]\), \(\sin^{-1}\) returns \(2\theta\): \[ 2\tan^{-1}x=2\theta=\sin^{-1}\!\left(\frac{2x}{1+x^2}\right). \]