Given $0 \le x \le \dfrac{1}{2}$, then the value of

$tan\left [ sin^{-1}\left\{\dfrac{x}{\sqrt{2}} + \dfrac{\sqrt{1-x^2}}{\sqrt{2}} \right \} - sin^{-1}x \right ]$

The correct answer is 1

$tan\left [ sin^{-1}\left\{\dfrac{x}{\sqrt{2}} + \dfrac{\sqrt{1-x^2}}{\sqrt{2}} \right \} - sin^{-1}x \right ]$

= $tan\left [ sin^{-1}\left\{\dfrac{x +\sqrt{1-x^2}}{\sqrt{2}} \right \} - sin^{-1}x \right ]$

Let $sin^{-1}x = \theta$ ⇒ $x = sin\theta$

⇒ $tan\left [ sin^{-1}\left\{\dfrac{sin\theta +\sqrt{1-sin^2\theta}}{\sqrt{2}} \right \} - \theta \right ]$

⇒ $tan\left [ sin^{-1}\left\{\dfrac{sin\theta +cos\theta}{\sqrt{2}} \right \} - \theta \right ]$

⇒ $tan\left [ sin^{-1}\left\{sin(\theta + \dfrac{\pi}{4}) \right \} - \theta \right ]$

⇒ $tan\left [ \theta + \dfrac{\pi}{4} - \theta \right ]$

⇒ $tan\left [ \dfrac{\pi}{4} \right ]$

⇒ 1