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Question

If sin1x+sin1y+sin1z=3π2
then 2r=1(x100r+y103r)x201y201 =

A
\N
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B
2
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C
4
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D
43
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Solution

The correct option is D 43
We know that π2sin1xπ2π2sin1yπ2π2sin1zπ2sin1x+sin1y+sin1z=3π2sin1x=sin1y=sin1z=π2x=y=z=12r=1(x100r+y103r)x201y201=x100+y103+x200+y206x201y201+y201z201+z201x201=1+1+1+11.1+1.1+1.1=43

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