A) \[-\frac{\pi }{2}\]
B) \[\frac{\pi }{2}\]
C) 0
D) None of these
Correct Answer: B
Solution :
\[\because \,\,{{\cos }^{-\,1}}\cos x={{\cos }^{-\,1}}\cos (2\pi -x)\] For \[x\in \left( \frac{3\pi }{2},2\pi \right)=2\pi -x\] |
\[{{\sin }^{-1}}\sin x=x-2\pi \] |
\[\therefore \,\,{{\sin }^{-\,1}}(\cos \,({{\cos }^{-\,1}}(\cos x)+{{\sin }^{-\,1}}(\sin x)))\] |
\[={{\sin }^{-1}}(\cos 0)={{\sin }^{-1}}(1)=\frac{\pi }{2}\] |
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