tanα and tanβ are roots of the equation x2+ax+b=0 , then the
value of sin2(α+β)+asin(α+β)⋅cos(α+β)+bcos2(α+β) is equal to
ab
b
a
tanα+tanβ=−a,tanα⋅tanβ=btan(α+β)=tanα+tanβ1-tanα⋅tanβ=-a1-b=ab-1⇒ sin2(α+β)+a⋅sin(α+β)cos(α+β)+bcos2(α+β)=cos2(α+β)tan2(α+β)+b+atan(α+β) =tan2(α+β)+b+atan(α+β)1+tan2(α+β) cos2α+β=1 sec2α+β=11+tan2(α+β)=ab−1a+ab−1+b1+a2(b−1)2=a2b-1+a2(b-1)2+b1+a2(b−1)2=a2(b-1)+a2(b-1)2+b1+a2(b−1)2a2bb-12+b1+a2(b−1)2=b