A) \[{{\cos }^{-1}}\sqrt{1-{{x}^{2}}}\]
B) \[-{{\cos }^{-1}}\sqrt{1-{{x}^{2}}}\]
C) \[{{\cos }^{-1}}\sqrt{{{x}^{2}}-1}\]
D) \[\pi -{{\cos }^{-1}}\sqrt{1-{{x}^{2}}}\]
Correct Answer: B
Solution :
Let \[{{\sin }^{-1}}x=y.\] Then \[x=\sin y\] Since \[-1\le x\le 0,\] therefore \[\frac{-\pi }{2}\le {{\sin }^{-1}}x\le 0\] and so \[\frac{-\pi }{2}\le y\le 0\] We have \[\cos y=\sqrt{1-{{\sin }^{2}}y}\] \[\Rightarrow \,\,\cos y=\sqrt{1-{{x}^{2}}}\], for \[0\le y\le \pi \] ?..(i) Now \[-\frac{\pi }{2}\le y\le 0\,\,\Rightarrow \,\,\frac{\pi }{2}\ge -y\ge 0\] \[\Rightarrow \,\,\cos \,\left( -y \right)=\sqrt{1-{{x}^{2}}}\] {from (i)} \[\Rightarrow \,\,-y={{\cos }^{-1}}\sqrt{1-{{x}^{2}}}\,\,\Rightarrow \,\,y=-{{\cos }^{-1}}\sqrt{1-{{x}^{2}}}\].
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