Let −π6<θ<−π12. Suppose α1 and β1 are the roots of the equation x2−2xsecθ+1=0 and α2 and β2 are the roots of the equation x2+2xtanθ−1=0. If α1>β1 and α2>β2, then α1+β2 equals
2(secθ−tanθ)
2secθ
−2tanθ
0
Solving given equations and from given values of θ which lie in fourth quadrant
α1=secθ−tanθβ1=secθ+tanθα2=secθ−tanθβ2=−secθ−tanθα1+β2=−2tanθ