Let A -1-sin- cos- - B -1-cos-sin- If - - are the roots of quadratic equation 3 sin-2 sin 2-1-0 where 0 -2 and the distance AB - a - b units- then a - b -

Given : 3sinθ 2sin^2θ 1=0 ⇒ (sinθ 1)(1 2sinθ)=0 ⇒ sinθ=1 or sinθ=12 ∴ θ=π2 or π6 Let α=π6,β=π2 or nbsp;α=π2,β=π6 nbsp;in both the cases distance is same as ...

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Byju's Answer

Standard XII

Mathematics

Distance Formula

Let A ≡1+sinα...

Question

Let $A \equiv (1 + sin α, cos α)$ ; $B \equiv (1 - cos β, - sin β)$ . If $α, β$ are the roots of quadratic equation $3 sin θ - 2 {sin}^{2} θ - 1 = 0$ where $0 \leq θ \leq \frac{π}{2}$ and the distance $A B = \sqrt{a + \sqrt{b}} units$ , then $a + b =$

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Solution

Given : $3 sin θ - 2 {sin}^{2} θ - 1 = 0$
$\Rightarrow (sin θ - 1) (1 - 2 sin θ) = 0$
$\Rightarrow sin θ = 1$ or $sin θ = \frac{1}{2}$
$∴ θ = \frac{π}{2} or \frac{π}{6}$
Let $α = \frac{π}{6}, β = \frac{π}{2}$ or $α = \frac{π}{2}, β = \frac{π}{6}$ in both the cases distance is same as
$A B = \sqrt{(sin α + cos β)^{2} + (cos α + sin β)^{2}}$
$= \sqrt{2 + 2 sin (α + β)} = \sqrt{2 + \sqrt{3}}$
comparing with $\sqrt{a + \sqrt{b}} \Rightarrow a = 2, b = 3$
$∴ a + b = 5$

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Let A≡(1+sinα,cosα); B≡(1−cosβ,−sinβ). If α,β are the roots of quadratic equation 3sinθ−2sin2θ−1=0 where 0≤θ≤π2 and the distance AB=√a+√b units, then a+b=

Given : 3sinθ−2sin2θ−1=0 ⇒ (sinθ−1)(1−2sinθ)=0 ⇒ sinθ=1 or sinθ=12 ∴ θ=π2 or π6 Let 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

Let $A \equiv (1 + sin α, cos α)$ ; $B \equiv (1 - cos β, - sin β)$ . If $α, β$ are the roots of quadratic equation $3 sin θ - 2 {sin}^{2} θ - 1 = 0$ where $0 \leq θ \leq \frac{π}{2}$ and the distance $A B = \sqrt{a + \sqrt{b}} units$ , then $a + b =$