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Let $0 \leq a, b, c, d \leq π$ and a,b,c,d are not complimentary such that $2 \cos a + 6 \cos b + 7 \cos c + 9 \cos d = 0$ and $2 \sin a - 6 \sin b + 7 \sin c - 9 \sin d = 0$ . Then the value of $\frac{\cos (a + d)}{\cos (b + c)} =$

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Correct option is C

We have(2cosa+9cosd)2=(6cosb+7cosc)2-----(1)(2sina-9sind)2=(6sinb-7sinc)2--------(2)Adding we get cosa+dcosb+c=8436=73

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Let0≤a,b,c,d≤π and a,b,c,d are not complimentary such that 2cosa+6cosb+7cosc+9cosd=0 and 2sina−6sinb+7sinc−9sind=0. Then the value of cosa+dcosb+c=

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Let $0 \leq a, b, c, d \leq π$ and a,b,c,d are not complimentary such that $2 \cos a + 6 \cos b + 7 \cos c + 9 \cos d = 0$ and $2 \sin a - 6 \sin b + 7 \sin c - 9 \sin d = 0$ . Then the value of $\frac{\cos (a + d)}{\cos (b + c)} =$