A) 1
B) \[-\frac{1}{\sqrt{2}}\]
C) \[\frac{1}{\sqrt{2}}\]
D) None of these
Correct Answer: C
Solution :
[c] \[{{\sin }^{-1}}(x-1)\Rightarrow -1\le x-1\le 1\Rightarrow 0\le x\le 2\] \[{{\cos }^{-1}}(x-3)\Rightarrow -1\le x-3\le 1\Rightarrow 2\le x\le 4\] \[\therefore x=2\] So, \[{{\sin }^{-1}}(2-1)+co{{s}^{-1}}(2-3)+ta{{n}^{-1}}\frac{2}{2-4}\] \[={{\cos }^{-1}}k+\pi \] Or \[{{\sin }^{-1}}1+{{\cos }^{-1}}(-1)+ta{{n}^{-1}}(-1)=co{{s}^{-1}}k+\pi \] \[\Rightarrow \frac{\pi }{2}+\pi -\frac{\pi }{4}={{\cos }^{-1}}k+\pi \Rightarrow {{\cos }^{-1}}k=\frac{\pi }{4}\]or \[k=\frac{1}{\sqrt{2}}\]
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