If `alphaa n dbeta` 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

Given,

`bsectheta=c−atantheta`

Squaring both sides,

`(bsectheta)^2=(c−atantheta)^2`

`b^2sec^2theta=c^2−a^2tan^2theta-2xatantheta`

`=>b^2(1+tan^2theta)=c^2−2catantheta+a^2tan^2theta`

`=>(a^2−b^2)tan^2theta−2catantheta+(c^2−a^2)=0`

comparing,

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