Let A and B denote the statements A: α + β + γ = 0 B: α + β + γ = 0 If (β-γ) + (γ-α) + (α-β) = -3/2, then: [AIEEE 2009]

Correct Answer is : both AAA and BBB are true

Correct Answer is : both AAA and BBB are true

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AIEEE 2009 Solved Paper

Section: Mathematics

Question : 60 of 89

Marks: +1, -0

Let

A

B

\cos α + \cos β + \cos γ = 0

\sin α + \sin β + \sin γ = 0

\cos (β-γ) + \cos (γ-α) + \cos (α-β) = -\frac{3}{2}

Solution:

We have

\cos (β-γ) + \cos (γ-α) + \cos (α-β) = -\frac{3}{2}

\Rightarrow 2[\cos (β-γ) + \cos (γ-α) + \cos (α-β)] + 3 = 0

\Rightarrow 2[\cos (β-γ) + \cos (γ-α) + \cos (α-β)] + \sin^{2} α + \cos^{2} α + \sin^{2} β + \cos^{2} β

+ \sin^{2} γ + \cos^{2} γ = 0

\Rightarrow 2[(\cos β \cos γ + \sin β \sin γ) + (\cos γ \cos α + \sin γ \sin α)

+ (\cos α \cos β + \sin α \sin β)] + + \sin^{2} α + \cos^{2} α

+ \sin^{2} β + \cos^{2} β + \sin^{2} γ + \cos^{2} γ = 0

\Rightarrow [ \sin^{2} α + \sin^{2} β + \sin^{2} γ + 2 \sin α \sin β + 2 \sin β \sin γ + 2 \sin γ \sin α] +

[ \cos^{2} α + \cos^{2} β + \cos^{2} γ + 2 \cos α \cos β + 2 \cos β \cos γ + 2 \cos γ \cos α] = 0

\Rightarrow[ \sin α + \sin β + \sin γ]^{2} + (\cos α + \cos β + \cos γ)^{2} = 0

\Rightarrow \sin α + \sin β + \sin γ = 0

and

\cos α + \cos β + \cos γ = 0

\therefore A

and

B

both are true.

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