text: • # question_answer Statement-1: If$1/2\le x\le 1$, then ${{\cos }^{-1}}x+{{\cos }^{-1}}\left[ \frac{x}{2}+\frac{\sqrt{3-3{{x}^{2}}}}{2} \right]$is equal to$\pi /3$ Statement 2:${{\sin }^{-1}}\left( 2x\sqrt{1-{{x}^{2}}} \right)=2{{\sin }^{-1}}x,$if$x\in \left[ -1/\sqrt{2},\,\,1/\sqrt{2} \right]$ A)  Statement-1 is true, Statement-2 is true; Statement-2 is a correct explanation for Statement-1.B)  Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1.C)  Statement-1 is true, Statement-2 is false.D)  Statement-1 is false, Statement-2 is true.
  
  In statement $1$ put $x=\cos \theta$ then$0\le \theta \le \pi /3$ $L.H.S={{\cos }^{-1}}(\cos \theta )+{{\cos }^{-1}}\left[ \frac{1}{2}\cos \theta +\frac{\sqrt{3}}{2}\sin \theta \right]$$=\theta +{{\cos }^{-1}}(\cos (\pi /3-\theta ))=\theta +\pi /3-\theta =\frac{\pi }{3}$so statement $1$ is true. In statement $2$, put $x=\sin \theta$ then$-\pi /4\le \theta \le \pi /4$ so${{\sin }^{-1}}(2x\sqrt{1-{{x}^{2}}})={{\sin }^{-1}}(2\sin \theta \cos \theta )$ $={{\sin }^{-1}}(\sin 2\theta )=2\theta =2{{\sin }^{-1}}x$