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 \[\therefore \] \[x\in \left( \frac{3\pi }{2},\,2\pi \right)\Rightarrow \,{{\cos }^{-1}}\left( \cos x \right)=2\pi -x\] \[\therefore \] \[{{\cos }^{-1}}(\cos x)+{{\sin }^{-1}}(\sin x)=x-2\pi \]\[+2\pi -x=0\] \[\therefore \] \[{{\sin }^{-1}}\left\{ \cos ({{\cos }^{-1}}(\cos x)+{{\sin }^{-1}}(\sin x)) \right\}\] \[={{\sin }^{-1}}\left\{ \cos \,0 \right\}={{\sin }^{-1}}1=\frac{\pi }{2}\]You need to login to perform this action.
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