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Q.

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

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a

0

b

1

c

2

d

3

answer is A.

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Detailed Solution

Put  x=y=z=1
then x100+y100+z1009x101+y101+z101=   1+1+191+1+1
                                                                                     =33=0

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