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$\begin{array}{l} I f \sin θ (1 + \sin θ) + \cos θ (1 + \cos θ) = x \\ a n d \sin θ (1 - \sin θ) + \cos θ (1 - \cos θ) = y t h e n \end{array}$

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x2−2x=sin2θ

y2+2y=sin2θ

xy=sin2θ

x−y=2

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detailed solution

Correct option is C

Givensinθ1+sinθ+cosθ1+cosθ=x⇒sinθ+sin2θ++cosθ+cos2θ=x⇒sinθ+cosθ=x-1and sinθ1−sinθ+cosθ1−cosθ=y⇒sinθ+cosθ=x−1=y+1∴x-y=2xy=sinθ+cosθ2-1=sin2θ

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If sinθ1+sinθ+cosθ1+cosθ=x and sinθ1−sinθ+cosθ1−cosθ=y then

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$\begin{array}{l} I f \sin θ (1 + \sin θ) + \cos θ (1 + \cos θ) = x \\ a n d \sin θ (1 - \sin θ) + \cos θ (1 - \cos θ) = y t h e n \end{array}$