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Question

In any ABC,(sin2A+sinA+1sinA) is always greater than

A
9
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B
3
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C
27
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D
none of these.
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Solution

The correct option is D 27
(sin2A+sinA+1sinA)
sinA+1sinA2 if A=900
sinB+1sinB>2
and sinC+1sinC>2
(sin2A+sinA+1sinA)
=(sinA+1+1sinA)(sinB+1+1sinB)(sinC+1+1sinC)
Re-arranged as
=(sinA+1sinA+1)(sinB+1sinB+1)(sinC+1sinC+1)
>(2+1)(2+1)(2+1)
>3×3×3=27
(sin2A+sinA+1sinA)>27

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