Question

( tan A+tan B=frac{10}{3} )
( tan A tan B=frac{-25}{3} )
( tan (A+B)=frac{tan A+tan B}{1-tan A tan B}=frac{10 / 3}{1+frac{25}{3}}=frac{5}{14} )
so sin ( (A+B)=frac{5}{sqrt{221}} ) and ( cos (A+B)=frac{14}{sqrt{227}} )
( therefore 3 sin ^{2}(A+B)-10 sin (A+B) cos (A+B)-25 cos ^{2}(A+B) )
( =3 times frac{25}{221}-frac{10 times 5 times 14}{221}-25 times frac{14^{2}}{221} )
( =frac{25}{221}(3-28-196)=-25 )

# 5 If tan A and tan B are the roots of the quadratic equation, 3x2 – 10x – 25 = 10 then the value of 3 sin(A + B) – 10sin(A + B). COS(A + B) – 25 cos2(A + B) is [JEE-Main Online-2018] (A) 25 (B) –25 (C) -10 (D) 10

Solution