Let  $0\le a,b,c,d\le \pi$ where b, c are not complementary 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  $\frac{\cos (a+d)}{\cos (b+c)}=$

$$\begin{gathered} {(2\cos a+9\cos d)}^2={(6\cos b+7\cos c)}^2 \\ 4{\cos }^{2}a+81{\cos }^{2}d+36\cos a\cos d \\ =36{\cos }^{2}b+49{\cos }^{2}c+84\cos b\cos c\mathrm{......}\left(1\right) \\ {(2\sin a-9\sin d)}^2={(6\sin b-7\sin c)}^2 \\ 4{\sin }^{2}a+81{\sin }^{2}d-36\sin a\sin d \\ =36{\sin }^{2}b+49{\sin }^{2}c-84\sin b\sin c\mathrm{..........}\left(2\right) \end{gathered}$$

Adding (1) and (2) 

$$\begin{gathered} 4+81+36\cos (a+d)=36+49+84\cos (b+c) \\ \frac{\cos (a+d)}{\cos (b+c)}=\frac{84}{36}=\frac{7}{3} \end{gathered}$$