A) \[\frac{1}{\sqrt{2}}\]
B) \[-\frac{1}{\sqrt{2}}\]
C) \[\frac{-\sqrt{3}}{2}\]
D) \[\frac{\sqrt{3}}{2}\]
E) \[\frac{1}{2}\]
Correct Answer: D
Solution :
\[2{{\sin }^{-1}}x-{{\cos }^{-1}}x=\frac{\pi }{2}\] \[\Rightarrow \]\[2{{\sin }^{-1}}x+2{{\cos }^{-1}}x-2{{\cos }^{-1}}x\] \[-{{\cos }^{-1}}x=\frac{\pi }{2}\] \[\Rightarrow \]\[2({{\sin }^{-1}}x+{{\cos }^{-1}}x)-3{{\cos }^{-1}}x=\frac{\pi }{2}\] \[\Rightarrow \]\[2.\frac{\pi }{2}-3{{\cos }^{-1}}x=\frac{\pi }{2}\] \[\left( \because {{\sin }^{-1}}x+{{\cos }^{-1}}x=\frac{\pi }{2} \right)\] \[\Rightarrow \] \[\pi -\frac{\pi }{2}-3{{\cos }^{-1}}x=\frac{\pi }{2}\] \[\Rightarrow \] \[\frac{\pi }{2}=3{{\cos }^{-1}}x\] \[\Rightarrow \] \[{{\cos }^{-1}}x=\frac{\pi }{6}\] \[\Rightarrow \] \[x=\cos \frac{\pi }{6}\] \[\Rightarrow \] \[x=\frac{\sqrt{3}}{2}\]
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