If tan A and tan B are the roots of the quadratic equation x^2-ax+b=0 then find the value of sin^2(A+B)?

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Answer 1 · 2018-06-03T15:58:41

$sin^2(A+B)=a^2/(a^2+b^2-2b+1)$

We know that,

$"If"$ $alpha and beta$ $"are the roots of the quadratic equation :"$

$color(red)(Px^2+Qx+R=0,then,alpha+beta=-Q/Pand alpha*beta=R/P$

We have,

$x^2-ax+b=0=>P=1 , Q=-a and R=b$

The roots are: $alpha=tanA and beta=tanB$

So,

$tanA+tanB=-Q/P=-(-a)/1=a$

$tanA*tanB=R/P=b/1=b$

Now,

$tan(A+B)=(tanA+tanB)/(1-tanAtanB)=a/(1-b)$

$=>tan^2(A+B)=a^2/(1-2b+b^2)$

#=>cot^2(A+B)=(1-2b+b^2)/a^2..to[becausecottheta=1/tantheta]#

$=>1+cot^2(A+B)=1+(1-2b+b^2)/a^2to$ [add. bothsides $1$ ]

$=>csc^2(A+B)=(a^2+1-2b+b^2)/a^2$

$=>sin^2(A+B)=a^2/(a^2+1-2b+b^2)..to[becausesintheta=1/(csctheta) ]$

Note :
We have take the quadratic equn. $Px^2+Qx+R=0$ ,because
$A, B, a, b,$ are used in the question.