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Question

If α,β,γ,δ are the smallest positive angles in ascending order of magnitude which have their sines equal to the positive quantity k, then the value of 4 sinα2+3 sinβ2+2 sinγ2+sin δ2 is equal to

A
21k
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B
121+k
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C
21+k
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D
None of these
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Solution

The correct option is C 21+k
Given α < β < γ < δ and sin α=sin β=sin γ=sinδ=k . Also α, β, γ, δ are smallest positive angles satisfying above two conditions.
We can take β=πα,γ=2π+α,δ=3πα
Given expression
​ ​=4 sin α2+3 sin (π2α2)+2 sin(π+α2)+sin(3π2α2)=4 sin α2+3 cos α22 sinα2cosα2=2(sinα2+cosα2)=2(sin12α+cos12α)2=21+sin α=21+k

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