If sin−1x+sin−1y+sin−1z=3π2 then the value of x100+y100+z100−3x101+y101+z101 is
0
1
2
3
We have sin–1x + sin–1y + sin–1z = 3π2 it is possible only whensin−1x=π2⇒x=1 ∵sin−1x≤π2sin−1y=π2⇒y=1; sin−1z=π2⇒z=1∴x100+y100+z100−3x101+y101+z101=1+1+1−33=3−1=2