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If $p = \sin (α - β) \sin (γ - δ), q = \sin (β - γ) \sin (α - δ)$ and $r = \sin (γ - α) \sin (β - δ)$ , then

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p + q + r = 0

p + q - r = 0

p - q - r = 0

p^{3} + q^{3} + r^{3} - 3 p q r = 0

answer is D, A.

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Detailed Solution

p = \sin (α - β) . \sin (γ - δ)

$q = \sin (β - γ) . \sin (α - δ)$ and

$r = \sin (γ - α) . \sin (β - δ)$

$\Rightarrow p + q + r = \sin (α - β) . \sin (γ - δ) + \sin (β - γ) . \sin (α - δ) + \sin (γ - α) . \sin (β - δ)$

$= \frac{1}{2} [\cos (α - β - γ + δ) - \cos (α - β + γ - δ) + \cos (β - γ - α + δ) - \cos (β - γ + α - δ)$

$+ \cos (γ - α - β + δ) - \cos (γ - α + β - δ)]$

$[∵ \cos (α - β - γ + δ) = \cos (γ - α + β - δ)$

$= \cos (- (α - β - γ + δ))$

$\cos (α - β - γ + δ) = \cos (α - β - γ + δ)]$

$∴ p + q + r = 0 \Rightarrow p^{3} + q^{3} + r^{3} = 3 p q r$

$p + q + r = 0$ , $p^{3} + q^{3} + r^{3} - 3 p q r = 0$

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