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Question

Suppose α,β and θ are angles satisfying 0<α<θ<β<π2, then sinαsinβcosβcosα=


A
tanθ
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B
tanθ
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C
cotθ
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D
cotθ
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Solution

The correct option is C cotθ
Let f(x)=sinx and g(x)=cosx, then f and g are continuous and derivable. Also, sinx0 for any x(0,π2)
So by Cauchy's mean value theorem, f(β)f(α)g(β)g(α)=f(θ)g(θ)
sinβsinαcosβcosα=cosθsinθ=cotθ
sinαsinβcosβcosα=cotθ

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