设 $ alpha+gamma=theta $ ,则有 $$ sin2theta=3sin2beta,$$ 令 $ x=theta+beta,y=theta-beta $ ,则 $$ 2theta=x+y,2beta=x-y,$$ 所以 $$ sinleft( x+y right)=3sinleft( x-y right)$$ 所以 $$ sin xcos y+cos xsin y=3sin xcos y-3cos xsin y,$$ 所以 $$ 4cos xsin y=2sin xcos y,$$ 所以 $ tan x=2tan y $ ,从而
$$ m=dfrac{tanleft( theta+beta right)}{tanleft( theta-beta right)}=dfrac{tan x}{tan y}=2.$$

设 $ alpha+gamma=theta $ ,则有 $$ sin2theta=3sin2beta,$$ 所以 $$ dfrac{2sinthetacostheta}{cos^2theta+sin^2theta}=dfrac{6sinbetacosbeta}{cos^2beta+sin^2beta},$$ 所以 $$ dfrac{tantheta}{1+tan^2theta}=dfrac{3tanbeta}{1+tan^2beta},$$ 令 $ x=tantheta,y=tanbeta $ ,上式可整理为 $$ x+xy^2=3y+3x^2y$$ 故 $$ begin{array}{rl} m=dfrac{tanleft( theta+beta right)}{tanleft( theta-beta right)}&=dfrac{tantheta+tanbeta}{1-tanthetatanbeta}timesdfrac{1+tanthetatanbeta}{tantheta-tanbeta}\ &=dfrac{x+y}{1-xy}timesdfrac{1+xy}{x-y}\ &=dfrac{x+xy^2+y+x^2y}{x+xy^2-y-x^2y}\ &=dfrac{3y+3x^2y+y+x^2y}{3y+3x^2y-y-x^2y}\ &=dfrac{1+x^2+3+3x^2}{3+3x^2-1-x^2}=dfrac{4x^2+4}{2x^2+2}=2. end{array} $$

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