Let A≡(1+sinα,cosα); B≡(1−cosβ,−sinβ). If sinα,sinβ are the roots of quadratic equation 3sinθ−2sin2θ−1=0 where 0≤θ≤π2 and the distance AB=√a+√bunits, then a+b=
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Solution
Given : 3sinθ−2sin2θ−1=0 ⇒(sinθ−1)(1−2sinθ)=0 ⇒sinθ=1 or sinθ=12 ⇒sinα=1,12 ⇒α=π6,β=π2 or α=π2,β=π6 in both the cases distance is same as AB=√(sinα+cosβ)2+(cosα+sinβ)2 =√2+2sin(α+β)=√2+√3
comparing with √a+√b⇒a=2,b=3 ∴a+b=5