lf x=cos α +i sin α, y=cosβ+isinβ, thenx-yx+y=

x=cos α+i sin α, y=cos β+i sin βx-yx+y=cos α+i sin α-cos β-i sin βcos α+i sin α+cos β+i sin β =cos α-cos β+isin α-sin βcos α+cos β+isin α+sin β =-2αnα+β2sinα-β2+i2cos α+β2 sin α-β22 cos α+β2cos α-β2+i2sinα+β2cosα-β2 =2i sinα-β2 cosα+β2+i sinα+β22cosα-β2 cosα+β2+i sinα+β2 =i tan α-β2

lf x=cos α +i sin α, y=cosβ+isinβ, thenx-yx+y=

x=cos α+i sin α, y=cos β+i sin βx-yx+y=cos α+i sin α-cos β-i sin βcos α+i sin α+cos β+i sin β =cos α-cos β+isin α-sin βcos α+cos β+isin α+sin β =-2αnα+β2sinα-β2+i2cos α+β2 sin α-β22 cos α+β2cos α-β2+i2sinα+β2cosα-β2 =2i sinα-β2 cosα+β2+i sinα+β22cosα-β2 cosα+β2+i sinα+β2 =i tan α-β2

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$lf x = \cos α + i \sin α, y = cosβ + isinβ, then \frac{x - y}{x + y} =$

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$i . \cot (\frac{α - β}{2})$

$i . \sin (\frac{α - β}{2})$

$i . \cos (\frac{α - β}{2})$

$i . \tan (\frac{α - β}{2})$

answer is D.

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Detailed Solution

$\begin{array}{rcl} x & = & \cos α + i \sin α, y = \cos β + i \sin β \\ \frac{x - y}{x + y} & = & \frac{\cos α + i \sin α - \cos β - i \sin β}{\cos α + i \sin α + \cos β + i \sin β} \\ = & \frac{(\cos α - \cos β) + i (\sin α - \sin β)}{(\cos α + \cos β) + i (\sin α + \sin β)} \\ = & \frac{- 2 αn (\frac{α + β}{2}) \sin (\frac{α - β}{2}) + i (2 \cos \frac{α + β}{2} \sin (\frac{α - β}{2}))}{2 \cos (\frac{α + β}{2}) \cos (\frac{α - β}{2}) + i (2 \sin (\frac{α + β}{2}) \cos (\frac{α - β}{2}))} \\ = & \frac{2 i \sin (\frac{α - β}{2}) (\cos (\frac{α + β}{2}) + i \sin (\frac{α + β}{(2)}))}{2 \cos (\frac{α - β}{2}) (\cos (\frac{α + β}{2} + i \sin (\frac{α + β}{2})))} \\ = & i \tan (\frac{α - β}{2}) \end{array}$