If A+B+C+D=180∘ then cos⁡Acos⁡B+cos⁡Ccos⁡D=

I=cos⁡Acos⁡B+cos⁡Ccos⁡D=12[2cos⁡Acos⁡B+2cos⁡Ccos⁡D]=12[cos⁡(A+B)+cos⁡(A−B)+cos⁡(C+D)+cos⁡(C−D)]I=12[cos⁡Acos⁡B+sin⁡Asin⁡B+cos⁡Ccos⁡D+sin⁡Csin⁡D]=12(I+sin⁡Asin⁡B+sin⁡CsinD)⇒I−I2=12(sin⁡Asin⁡B+sin⁡Csin⁡D)⇒I=sin⁡Asin⁡B+sin⁡Csin⁡D

If A+B+C+D=180∘ then cos⁡Acos⁡B+cos⁡Ccos⁡D=

I=cos⁡Acos⁡B+cos⁡Ccos⁡D=12[2cos⁡Acos⁡B+2cos⁡Ccos⁡D]=12[cos⁡(A+B)+cos⁡(A−B)+cos⁡(C+D)+cos⁡(C−D)]I=12[cos⁡Acos⁡B+sin⁡Asin⁡B+cos⁡Ccos⁡D+sin⁡Csin⁡D]=12(I+sin⁡Asin⁡B+sin⁡CsinD)⇒I−I2=12(sin⁡Asin⁡B+sin⁡Csin⁡D)⇒I=sin⁡Asin⁡B+sin⁡Csin⁡D

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$If A + B + C + D = 180^{\circ} then \cos Acos B + \cos Ccos D =$

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sin A sin B + sin C sin D

sin A sin B - sin C sin D

sin A + sin B + sin C sin D

sin A sin B sin C sin D

answer is A.

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Detailed Solution

$\begin{array}{l} I = \cos Acos B + \cos Ccos D = \frac{1}{2} [2 \cos Acos B + 2 \cos Ccos D] \\ = \frac{1}{2} [\cos (A + B) + \cos (A - B) + \cos (C + D) + \cos (C - D)] \\ I = \frac{1}{2} [\cos Acos B + \sin Asin B + \cos Ccos D + \sin Csin D] = \frac{1}{2} (I + \sin Asin B + \sin CsinD) \\ \Rightarrow I - (\frac{I}{2}) = \frac{1}{2} (\sin Asin B + \sin Csin D) \Rightarrow I = \sin Asin B + \sin Csin D \end{array}$