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Question

If α,β are the roots of the equation x22x+2=0 and if cotθ=x+1, then [(x+α)n(x+β)n][αβ] is equal to

A
sinnθsinnθ
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B
cosnθcosnθ
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C
sinnθcosnθ
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D
cosnθsinnθ
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Solution

The correct option is A sinnθsinnθ
x22x+2=0&cotθ=x+1
α and β are the roots of the equation.
α,β=1+i,1i
x+α=cotθ+i
x+α=cosecθ(cis(θ))
(x+α)n=(cosecθ)n(cis(nθ))
Similarly,
(x+β)n=(cosecθ)n(cis(nθ))
Now,
(x+α)n(x+β)n
=(cosecθ)n(cis(nθ)cis(nθ))
=(cosecθ)n(2isin(nθ))
=2isin(nθ)(sinθ)n
Also, αβ=2.
Hence, Given expression is 2isin(nθ)(sinθ)n2i, which is =sin(nθ)(sinθ)n
Hence, option A is correct.

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