If α,β are the roots of the equation x2−2x+2=0 and if cotθ=x+1, then [(x+α)n−(x+β)n][α−β] 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 Asinnθsinnθ x2−2x+2=0&cotθ=x+1 α and β are the roots of the equation. ⇒α,β=1+i,1−i 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, α−β=2.
Hence, Given expression is 2isin(nθ)(sinθ)n2i, which is =sin(nθ)(sinθ)n Hence, option A is correct.