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Q.

If  sin1x+sin1y+sin1z=3π2, then  the value of x100+y100+z1009x101+y101+z101 is

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a

1

b

0

c

-1

d

3

answer is C.

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Detailed Solution

Givensin1x+sin1y+sin1z=3π2

let​  ​sin1x=sin1y=sin1z=π2x=y=z=1x100+y100+z1009x101+y101+z101=0

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If  sin−1x+sin−1y+sin−1z=3π2, then  the value of x100+y100+z100−9x101+y101+z101 is