If 32+cos⁡θ+isin⁡θ=x+iy then (x−1)(x−3)=

Given : 32+cos⁡θ+isin⁡θ=x+iyOn Rationalizing , we get ⇒x+iy=32+cos⁡θ+isin⁡θ×2+cos⁡θ-isin⁡θ2+cos⁡θ-isin⁡θ ⇒x+iy=6+3cos⁡θ-3i sin⁡θ2+cos⁡θ2-i2sin⁡θ ⇒x+iy=6+3cos⁡θ5+4cosθ+-3i sin⁡θ5+4cosθOn comparing we get ⇒x=6+3cos⁡θ5+4cosθ and y=-3 sin⁡θ5+4cosθ⇒(x−1)(x−3)=6+3cos⁡θ5+4cosθ-16+3cos⁡θ5+4cosθ-3 ⇒(x−1)(x−3)=-91-cos⁡θ5+4cosθ1-cos⁡θ5+4cosθ ⇒(x−1)(x−3)=-9 sin2⁡θ5+4cosθ2=-y2

If 32+cos⁡θ+isin⁡θ=x+iy then (x−1)(x−3)=

Given : 32+cos⁡θ+isin⁡θ=x+iyOn Rationalizing , we get ⇒x+iy=32+cos⁡θ+isin⁡θ×2+cos⁡θ-isin⁡θ2+cos⁡θ-isin⁡θ ⇒x+iy=6+3cos⁡θ-3i sin⁡θ2+cos⁡θ2-i2sin⁡θ ⇒x+iy=6+3cos⁡θ5+4cosθ+-3i sin⁡θ5+4cosθOn comparing we get ⇒x=6+3cos⁡θ5+4cosθ and y=-3 sin⁡θ5+4cosθ⇒(x−1)(x−3)=6+3cos⁡θ5+4cosθ-16+3cos⁡θ5+4cosθ-3 ⇒(x−1)(x−3)=-91-cos⁡θ5+4cosθ1-cos⁡θ5+4cosθ ⇒(x−1)(x−3)=-9 sin2⁡θ5+4cosθ2=-y2

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If $\frac{3}{2 + \cos θ + isin θ} = x + iy$ then $(x - 1) (x - 3) =$

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$y^{2}$

$- y^{2}$

answer is B.

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Detailed Solution

Given : $\frac{3}{2 + \cos θ + isin θ} = x + iy$

On Rationalizing , we get

$\Rightarrow x + i y = \frac{3}{(2 + \cos θ) + isin θ} \times \frac{(2 + \cos θ) - isin θ}{(2 + \cos θ) - isin θ} \Rightarrow x + i y = \frac{6 + 3 \cos θ - 3 i \sin θ}{{(2 + \cos θ)}^{2} - i^{2} \sin θ} \Rightarrow x + i y = \frac{6 + 3 \cos θ}{5 + 4 \cos (θ)} + \frac{- 3 i \sin θ}{5 + 4 \cos (θ)}$

On comparing we get $\Rightarrow x = \frac{6 + 3 \cos θ}{5 + 4 \cos (θ)} a n d y = \frac{- 3 \sin θ}{5 + 4 \cos (θ)}$

$\Rightarrow (x - 1) (x - 3) = (\frac{6 + 3 \cos θ}{5 + 4 \cos (θ)} - 1) (\frac{6 + 3 \cos θ}{5 + 4 \cos (θ)} - 3) \Rightarrow (x - 1) (x - 3) = (\frac{- 9 (1 - \cos θ)}{5 + 4 \cos (θ)}) (\frac{1 - \cos θ}{5 + 4 \cos (θ)}) \Rightarrow (x - 1) (x - 3) = \frac{- 9 \sin^{2} θ}{{(5 + 4 \cos (θ))}^{2}} = - y^{2}$