If `alpha and beta ` are the roots of the equation `x^(2) - p(x + 1) - q = 0` , then the value of :
`(alpha^(2) + 2 alpha + 1)/(alpha^(2)+ 2 alpha + q) + (beta^(2) + 2 beta + 1)/(beta^(2) + 2 beta + q)` is :

Explanation

Since `alpha` and `beta` are the roots of the equation
`x^(2)-px-p-c=0`
`:.alpha+beta=p`
and `alpha beta=-p-c`
Now `(alpha+1)(beta+1)=alpha beta+alpha +beta+1`
`=-p-c+p+q=1-c`
Hence `(alpha+1)(beta+1)=1-c`……….i
Second part LHS`=(alpha^(2)+2alpha+1)/(alpha^(2)+2alpha+c)+(beta^(2)+2beta+1)/(beta^(2)+2beta+c)`
`=((alpha+1)^(2))/((alpha+1)^(2)-(1-c))+((beta+1)^(2))/((beta+1)^(2)-(1-c))`
`=(alpha+1)^(2))/((alpha+1)^(2)-(alpha+1)(beta+1))`
`+((beta+1)^(2))/((beta+1)^(2)-(alpha+1)(beta+1))` [from eq i ]
`=(alpha+1)/(alpha-beta)+(beta+1)/(beta-alpha)=(alpha-beta)/(alpha-beta)=-1=RHS`
Hence RHS=LHS`