Let sinθ-αsinθ-β=ab and cosθ-αcosθ-β=cd thenac+bdad+bc

sinθ-αsinθ-β=ab , cosθ-αcosθ-β=cdac+bdad+bc=2sin(θ-α).cos(θ-α)+2sin(θ-β)cos(θ-β)2sin(θ-α).cos(θ-β)+2sin(θ-β)cos(θ-α) =sin2(θ-α)+sin2(θ-β)sin(2θ-(α+β)+sin(β-α)+sin(2θ-(α+β))+sin(α-β) =2sin(2θ-(α+β)) cos(α-β)2sin(2θ-(α+β)) =cos (α-β)

Let sinθ-αsinθ-β=ab and cosθ-αcosθ-β=cd thenac+bdad+bc

sinθ-αsinθ-β=ab , cosθ-αcosθ-β=cdac+bdad+bc=2sin(θ-α).cos(θ-α)+2sin(θ-β)cos(θ-β)2sin(θ-α).cos(θ-β)+2sin(θ-β)cos(θ-α) =sin2(θ-α)+sin2(θ-β)sin(2θ-(α+β)+sin(β-α)+sin(2θ-(α+β))+sin(α-β) =2sin(2θ-(α+β)) cos(α-β)2sin(2θ-(α+β)) =cos (α-β)

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$Let \frac{\sin (θ - α)}{\sin (θ - β)} = \frac{a}{b}$ and $\frac{\cos (θ - α)}{\cos (θ - β)} = \frac{c}{d}$ $t h e n \frac{ac + bd}{ad + bc}$

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$\cos (α - β)$

$\sin (α - β)$

$\sin (α + β)$

$\cos (α + β)$

answer is A.

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Detailed Solution

$\frac{\sin (θ - α)}{\sin (θ - β)} = \frac{a}{b}$ , $\frac{\cos (θ - α)}{\cos (θ - β)} = \frac{c}{d}$

$\frac{ac + bd}{ad + bc}$

$= \frac{2 \sin (θ - α) . \cos (θ - α) + 2 \sin (θ - β) \cos (θ - β)}{2 \sin (θ - α) . \cos (θ - β) + 2 \sin (θ - β) \cos (θ - α)} = \frac{\sin 2 (θ - α) + \sin 2 (θ - β)}{\sin (2 θ - (α + β) + \sin (β - α) + \sin (2 θ - (α + β)) + \sin (α - β)} = \frac{2 \sin (2 θ - (α + β)) \cos (α - β)}{2 \sin (2 θ - (α + β))} = \cos (α - β)$