Question

If \[ \sin^{-1}(2x)-2\cos^{-1}\sqrt{1-x^2}=\frac{\pi}{2}, \] then \(\tan^{-1}(2x+1)=\)

• A. \(\dfrac{\pi}{6}\)
• B. \(\sin^{-1}x\)
• C. \(\dfrac{\pi}{3}\)
• D. \(\cos^{-1}x\)

Hint:

Useful identities: \[ \cos^{-1}\sqrt{1-x^2}=\sin^{-1}x, \] and \[ \tan^{-1}(\sqrt3)=\frac{\pi}{3}. \]

Answer 1

The Correct Option is C

Step 1: Simplify the inverse trigonometric expression. Using the identity \[ \cos^{-1}\sqrt{1-x^2}=\sin^{-1}x, \] the given equation becomes \[ \sin^{-1}(2x)-2\sin^{-1}x=\frac{\pi}{2}. \]

Step 2: Let \(\sin^{-1}x=\theta\). Then \[ x=\sin\theta \] and \[ \sin^{-1}(2x)=2\theta+\frac{\pi}{2}. \] Taking sine on both sides, \[ 2x=\sin\left(2\theta+\frac{\pi}{2}\right) =\cos2\theta. \] Since \[ \cos2\theta=1-2\sin^2\theta, \] we get \[ 2x=1-2x^2. \] Hence, \[ 2x^2+2x-1=0. \]

Step 3: Find the admissible value of \(x\). Solving, \[ x=\frac{-1\pm\sqrt3}{2}. \] Since \(2x\in[-1,1]\), \[ x=\frac{\sqrt3-1}{2}. \]

Step 4: Evaluate \(\tan^{-1}(2x+1)\). Now, \[ 2x+1 =2\left(\frac{\sqrt3-1}{2}\right)+1 =\sqrt3. \] Therefore, \[ \tan^{-1}(2x+1) =\tan^{-1}(\sqrt3) =\frac{\pi}{3}. \] Hence, \[ \boxed{\tan^{-1}(2x+1)=\frac{\pi}{3}}. \]