If α,β are solutions of sin2x+asinx+b=0 and cos2x+c cosx+d=0 , then sinα+β is equal to
2aca2+c2
a2+c22ac
2bdb2+d2
b2+d22bd
We have sinα+sinβ=-a, cosα+cosβ=-c ⇒sinα+sinβcosα+cosβ=ac ⇒2sinα+β2cosα-β22cosα+β2cosα-β2=ac ⇒tanα+β2=ac Now sinα+β=2tanα+β21+tan2α+β2=2aca2+c2