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`

If `acos2theta+bsin2theta=ch a salphaa n dbeta`as its roots , then pro

1.	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`
Answer» 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)`