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Name: JEE Advanced Chemistry
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Question

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Prashant Kotian · Accepted Answer

Correct answer: 5. Let xi = sin I [I = 1, 2, …….,10]     10 10 3 3 i i i 1 i 1 sin 0 4 sin 0          10 10 i i i 1 i 1 3 sin sin3 0        10 10 i i i 1 i 1 1 10 sin sin3 3 3      

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