If x+1x=2cosα, y+1y=2cosβ, then xy+1xy=

x+1x=2cos α ⇒ x2-2x cos α+1=0 ∴ x=2cos α±4cos2α-42=cos α±i sin αsimilarly y=cos β±i sin β ∴ xy+1xy =cos α+i sin α cos β±i sin β+1cos α±i sin α cos β±i sin β =cosα±β+i sin α±β+cosα±β-i sinα±β =2cos α±β

If x+1x=2cosα, y+1y=2cosβ, then xy+1xy=

x+1x=2cos α ⇒ x2-2x cos α+1=0 ∴ x=2cos α±4cos2α-42=cos α±i sin αsimilarly y=cos β±i sin β ∴ xy+1xy =cos α+i sin α cos β±i sin β+1cos α±i sin α cos β±i sin β =cosα±β+i sin α±β+cosα±β-i sinα±β =2cos α±β

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$If x + \frac{1}{x} = 2 cosα, y + \frac{1}{y} = 2 cosβ, then xy + \frac{1}{xy} =$

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$\cos (α \pm β)$

$2 \cos (α \pm β)$

$2 \sin (α \pm β)$

$\sin (α \pm β)$

answer is B.

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Detailed Solution

$\begin{array}{rcl} x + \frac{1}{x} & = & 2 \cos α \Rightarrow x^{2} - 2 x \cos α + 1 = 0 \end{array}$

$\begin{array}{rcl} ∴ x & = & \frac{2 \cos α \pm \sqrt{4 \cos^{2} α - 4}}{2} = \cos α \pm i \sin α \end{array}$

$\begin{array}{rcl} similarly y & = & \cos β \pm i \sin β \end{array}$

$\begin{array}{rcl} ∴ xy + \frac{1}{xy} \\ = & (\cos α + i \sin α) (\cos β \pm i \sin β) + \frac{1}{(\cos α \pm i \sin α) (\cos β \pm i \sin β)} \end{array}$