A) 2
B) 4
C) 6
D) 8
Correct Answer: A
Solution :
\[({{\sin }^{-1}}x)\in \,\left[ \frac{-\pi }{2},\,\frac{\pi }{2} \right]\] \[\therefore \,\,\,{{({{\sin }^{-1}}x)}^{2}}\,\le \frac{{{\pi }^{2}}}{4}\,\Rightarrow \,{{({{\sin }^{-1}})}^{2}}\,=\frac{{{\pi }^{2}}}{4}\] \[\therefore \,\,\,\,{{({{\sec }^{-1}}\,y)}^{2}}\,+{{({{\tan }^{-1}}\,z)}^{2}}\,=0\] or \[\,{{\sec }^{-1}}\,y={{\tan }^{-1}}\,z=0\] \[{{\sin }^{-1}}x=\pm \,\frac{\pi }{2}\,\Rightarrow \,x=\pm \,1,\,\,y=1,\,\,z=0\] \[\therefore \,\,(1,\,\,1,\,\,0)\,\,\And \,\,(-1,\,\,1,\,\,0)\]
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