Given tan A and tan B are the roots of `x^2-ax + b = 0`, 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. `(a^(2))/(b^(2)+(1-a)^(2))`

Answer

A

Explanation

Since tan A and tan B are the roots of the equation `x^(2)-ax+b=0.` Therefore,
`tanA+tanB=aand tan A tanB=b`
`thereforetan(A+B)=(a)/(1-b)`
Now, `sin^(2)(A+B)=1/2[1-cos2(A+B)]`
`impliessin^(2)(A+B)=1/2{1-(1-tan^(2)(A+B))/(1+tan^(2)(A+B))}`
`impliessin^(2)(A+B)=1/2{1-(1-(a^(2))/((1-b)^(2)))/(1+(a^(2))/((1-b)^(2)))}`
`impliessin^(2)(A+B)=1/2{(2a^(2))/(a^(2)+(1-b)^(2))}=(a^(2))/(a^(2)+(1-b)^(2))`