If `tan A & tan B` are the roots of the quadratic equation `x^2 - ax + b = 0`, then the value of `sin^2 (A +B)` is:

Options

A. `(a^(2))/(a^(2)+(1-b)^(2))`
B. `(a^(2))/(a^(2)+b^(2))`
C. `(a^(2))/((a+b)^(2))`
D. `(b^(2))/(a^(2)+(a-b)^(2))`

Answer

A

Explanation

`tanA+tanB=aand tanAtanB=b`
`tan(A+B)=(tanA+tanB)/(1-tanAtanB)=(a)/(1-b)`
Now , `sin^(2)(A+B)=1//2 " " [1cos2(A+B)]`
`=(1)/(2)[1-(1-tan^(2)(A+B))/(1+tan^(2)(A+B))]`
`=[(tan^(2)(A+B))/(1+tan^(2)(A+B))]`
`=(a^(2)//(1-b)^(2))/([(a^(2)+(1-b)^(2))/((1-b)^(2))])=(a^(2))/(a^(2)+(1-b)^(2))`