sin−1x+sin−1y+sin−1z=3π2, then the value of x100+y100+z100−9x101+y101+z101=
0
1
2
13
sin−1x+sin−1y+sin−1z=3π2 We know that ∣∣sin−1x∣∣≤π2 sin−1x=sin−1y=sin−1z=π2 x=y=z=sinπ2=1 x100+y100+z100−9x101+y101+z101=3−93=0