If `x/("tan"(theta+alpha))=y/("tan"(theta+beta))=z/(tan(theta+gamma)),` then show that `sum(x+y)/(x-y)sin^2(alpha-beta)=0`

Explanation

`(x)/(y)=(tan (theta+alpha))/(tan(theta+beta))`
By componendo and dividendo, we get
`(x+y)/(x-y)=(tan(theta+alpha)+tan(theta+beta))/(tan(theta+alpha)-tan(theta-beta))=(sin (2theta+alpha+beta))/(sin(alpha-beta))`
`therefore (x+y)/(x-y)sin^(2)(alpha-beta)=sin(2theta+alpha+beta)sin(alpha-beta)`
`=(1)/(2)[cos2(theta+beta)-cos2(theta+alpha)]`
Similarly,
`(y+z)/(y-z)sin^(2)(beta-gamma)=(1)/(2)[cos2(theta+gamma)-cos2(theta+1)`
and `(z+x)/(z-x)sin^(2)(gamma-alpha)=(1)/(2)(cos 2 (theta+alpha)-cos2(theta+gamma)`
Adding Eqs. i , ii, and iii, we get LHS=0.