If `alpha a n d beta` are the solutions of the equation `at a ntheta+bs e ctheta=c ,` then show that `tan(alpha+beta)=(2a c)/(a^2-c^2)`

Explanation

`atantheta+bsectheta=c`
`rArr bsectheta=c-a tan theta`
`rArr b^(2)sectheta=(c-a tantheta)^(2)`
`rArr b^(2)(1+tan^(2)theta=c^(2)+a^(2)tantheta-2ac tantheta`
`rArr tan^(2)theta(b^(2)-a^(2))+2actantheta + (b^(2)-c^(2))=0`
Now, the roots of the equation be `tanalpha` and `tanbeta`,
`therefore tanalpha + tanbeta=(-2ac)/(b^(2)-a^(2))`
and `tanalpha. tanbeta=(tanalpha+tanbeta)/(1-tanalphatanbeta)`
`=((-2ac)/(b^(2)-a^(2)))/(1-(b^(2)-c^(2))/(b^(2)-a^(2)))=(-2ac)/((b^(2)-a^(2))-(b^(2)-c^(2)))`