A) \[-\frac{\pi }{2}\]
B) \[\frac{\pi }{2}\]
C) 0
D) \[\pi \]
Correct Answer: B
Solution :
\[\because \] \[x\in \left( \frac{3\pi }{2},2\pi \right)\] Now, \[{{\cos }^{-1}}(\cos x)=2\pi -x\] and \[{{\sin }^{-1}}(\sin x)=x-2\pi \] \[\therefore \] \[{{\cos }^{-1}}(\cos x)+{{\sin }^{-1}}(\sin x)=0\] Therefore, \[{{\sin }^{-1}}[\cos \{{{\cos }^{-1}}(\cos x)+{{\sin }^{-1}}(\sin x)\}]\] \[={{\sin }^{-1}}\{cos(0)\}=si{{n}^{-1}}(1)=\frac{\pi }{2}\]You need to login to perform this action.
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