1.

If `tanbeta=(tanalpha+tangamma)/(1+tanalphatangamma)dot`prove that `sin2beta=(sin2alpha+sin2gamma)/(1+sin2alphasin2gamma)`.

Answer» `tanbeta = (tanalpha+tan gamma)/(1+tanalphatangamma)`
`=>tan beta = (sinalpha/cosalpha+singamma/cosgamma)/(1+sinalpha/cosalpha*singamma/cosgamma)`
`=(sinalphacosgamma + singammacosalpha)/(cosalphacosgamma+sinalphasingamma)`
`=>tan beta=(sin(alpha+gamma))/(cos(alpha-gamma))`
Now, `sin2beta = (2tanbeta)/(1+tan^2beta)`
`=(2(sin(alpha+gamma))/cos(alpha-gamma))/(1+sin^2(alpha+gamma)/cos^2(alpha-gamma) `
`=(2sin(alpha+gamma)cos(alpha-gamma))/(cos^2(alpha-gamma) + sin^2(alpha+gamma))`
`=(sin(alpha+gamma+alpha-gamma) + sin(alpha+gamma-alpha+gamma))/(((1+cos2(alpha-gamma))/2)+((1-cos2(alpha+gamma))/2))`
`=(sin2alpha+sin2gamma)/(1+1/2(cos(alpha-gamma) - cos(alpha+gamma)))`
`=(sin2alpha+sin2gamma)/(1+1/2(2sin2alphasin2gamma)`
`=(sin2alpha+sin2gamma)/(1+sin2alphasin2gamma) = R.H.S.`


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