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 4sinα2+3sinβ2+2sinγ2+sinδ2 is equal to
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a
21−k
b
21+k
c
1+k2
d
none of these
answer is B.
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Detailed Solution
Since α<β<γ<δ and sinα=sinβ=sinγ=sinδ=k, we have β=π−α,γ=2π+α,δ=3π−α⇒ 4sinα2+3sinβ2+2sinγ2+sinδ2 =4sinα2+3cosα2−2sinα2−cosα2 =2sinα2+2cosα2=21+sinα=21+k