Let sin(θ−α)sin(θ−β)=ab and cos(θ−α)cos(θ−β)=cd, then ac+bcdad+bc

Let a=k1sin(θ−α), b=k1sin(θ−β)And c=k2cos(θ−α), d=k2cos(θ−β)∴ac+bdad+bc=k1k2k1k2⋅sin(θ−α)⋅cos(θ−α)+sin(θ−β)⋅cos(θ−β)sin(θ−α)⋅cos(θ−β)+sin(θ−β)⋅cos(θ−α)∴ac+bdad+bc=k1k2k1k2⋅sin(θ−α)⋅cos(θ−α)+sin(θ−β)⋅cos(θ−β)sin(θ−α)⋅cos(θ−β)+sin(θ−β)⋅cos(θ−α)=sin2(θ−α)+sin2(θ−β)2sin(2θ−(α+β))=2sin(2θ−(α+β))⋅cos(α−β)2sin(2θ−(α+β))=cos(α−β).

Let sin(θ−α)sin(θ−β)=ab and cos(θ−α)cos(θ−β)=cd, then ac+bcdad+bc

Let a=k1sin(θ−α), b=k1sin(θ−β)And c=k2cos(θ−α), d=k2cos(θ−β)∴ac+bdad+bc=k1k2k1k2⋅sin(θ−α)⋅cos(θ−α)+sin(θ−β)⋅cos(θ−β)sin(θ−α)⋅cos(θ−β)+sin(θ−β)⋅cos(θ−α)∴ac+bdad+bc=k1k2k1k2⋅sin(θ−α)⋅cos(θ−α)+sin(θ−β)⋅cos(θ−β)sin(θ−α)⋅cos(θ−β)+sin(θ−β)⋅cos(θ−α)=sin2(θ−α)+sin2(θ−β)2sin(2θ−(α+β))=2sin(2θ−(α+β))⋅cos(α−β)2sin(2θ−(α+β))=cos(α−β).

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

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$\sin (α + β)$

$\sin (α - β)$

$\cos (α + β)$

$\cos (α - β)$

answer is A.

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Detailed Solution

Let $a = k_{1} \sin (θ - α), b = k_{1} \sin (θ - β)$

And $c = k_{2} \cos (θ - α), d = k_{2} \cos (θ - β)$

$∴ \frac{a c + b d}{a d + b c} = \frac{k_{1} k_{2}}{k_{1} k_{2}} \cdot \frac{\sin (θ - α) \cdot \cos (θ - α) + \sin (θ - β) \cdot \cos (θ - β)}{\sin (θ - α) \cdot \cos (θ - β) + \sin (θ - β) \cdot \cos (θ - α)}$