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Question

If α[π2,π], then the value of 1+sinα1sinα is

A
2cosα2
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B
2sinα2
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C
2
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D
2tanα2
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Solution

The correct option is A 2cosα2
1+sinα1sinα=(sinα2+cosα2)2(sinα2cosα2)2=sinα2+cosα2sinα2cosα2 (1)

Now, α[π2,π]
α2[π4,π2]sinα2>cosα2sinα2cosα2>0 (2)

From (1) and (2),
1+sinα1sinα=sinα2+cosα2sinα2+cosα2=2cosα2

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