If tanA and tanB are the roots of 3x2−10x−25=0, then the value of 3sin2(A+B)−10sin(A+B)cos(A+B)−25cos2(A+B) is
Solutions · 1
Official solution
tan(A+B)=1−tanAtanBtanA+tanB=1+25/310/3=145.
\[3\sin^2\phi-10\sin\phi\cos\phi-25\cos^2\phi=\cos^2\phi[3\tan^2\phi-10\tan\phi-25],\quad \phi=A+B.\]
Substituting tanϕ=145 and cos2ϕ=221196 gives the value −25.