question_answer

If \[x\in \left( \frac{3\pi }{2},\,2\pi  \right),\] then value of the expression \[{{\sin }^{-1}}\left\{ \cos ({{\cos }^{-1}}(\cos x)+{{\sin }^{-1}}(\sin x)) \right\}\] is equal to

A)  \[-\frac{\pi }{2}\]                                            

B)  \[\frac{\pi }{2}\]

C)  0                                            

D)  \[\frac{\pi }{2}-2x\]

Correct Answer: B

Solution :

The graph of \[y={{\sin }^{-1}}(\sin x)\] is \[\therefore \] If \[x\in \left( \frac{3\pi }{2},\,2\pi  \right)\Rightarrow \,{{\sin }^{-1}}(\sin x)=x-2\pi \] Again the graph of \[y={{\cos }^{-1}}(\cos x)\] is \[\therefore \] \[x\in \left( \frac{3\pi }{2},\,2\pi  \right)\Rightarrow \,{{\cos }^{-1}}\left( \cos x \right)=2\pi -x\] \[\therefore \] \[{{\cos }^{-1}}(\cos x)+{{\sin }^{-1}}(\sin x)=x-2\pi \]\[+2\pi -x=0\] \[\therefore \] \[{{\sin }^{-1}}\left\{ \cos ({{\cos }^{-1}}(\cos x)+{{\sin }^{-1}}(\sin x)) \right\}\] \[={{\sin }^{-1}}\left\{ \cos \,0 \right\}={{\sin }^{-1}}1=\frac{\pi }{2}\]