If `acos2theta+bsin2theta=ch a salphaa n dbeta` as its roots , then prove that `tanalpha+t a nbeta=(2a b)/(a+c)` `tanalpha+t a nbeta=(c-a)/(c+a)` `tan(alpha+beta)=b/a`

Explanation

Here, `acos2theta+bsin2theta = c`

`=>a((1-tan^2theta)/(1+tan^2theta)) +b((2tantheta)/(1+tan^2theta)) = c`

`=>a(1-tan^2theta)+2btantheta = c(1+tan^2theta)`

`=>(c+a)tan^2theta-2btantheta+(c-a) = 0->(1)`

Now, given equation has roots `alpha` and `beta`.

So, equation (1), will have roots `tan alpha` and `tan beta`.

`:.` Sum of the roots `= (-(-2b))/(c+a)`

`:. tanalpha+tanbeta = (2b)/(a+c)`