A) \[-{{\cos }^{-1}}\sqrt{1-{{x}^{2}}}\]
B) \[{{\cos }^{-1}}\sqrt{{{x}^{2}}-1}\]
C) \[\pi -{{\cos }^{-1}}\sqrt{1-{{x}^{2}}}\]
D) \[{{\cos }^{-1}}\sqrt{1-{{x}^{2}}}\]
Correct Answer: A
Solution :
Let \[{{\sin }^{-1}}x=y.\] Then, \[x=\sin y\] Since, \[-1\le x<0,\] therefore \[-\frac{\pi }{2}\le {{\sin }^{-1}}x<0\] \[\Rightarrow \] \[-\frac{\pi }{2}\le y<0\] Now, \[\cos \,y=\sqrt{1-{{x}^{2}}}\] for \[0\le y\le \pi \] But \[-\frac{\pi }{2}\le y<0\] \[\Rightarrow \] \[\frac{\pi }{2}\ge -y>0\] \[\Rightarrow \] \[\cos (-y)=\sqrt{1-{{x}^{2}}}\] \[\Rightarrow \] \[-y={{\cos }^{-1}}\sqrt{1-{{x}^{2}}}\] \[\Rightarrow \] \[y=-{{\cos }^{-1}}\sqrt{1-{{x}^{2}}}\]
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