Question 83
Let $a$, $b$, $c$ and $d$ be real numbers such that
$$
2 \cos a + 6 \cos b + 7 \cos c + 9 \cos d = 0, \\
2 \sin a - 6 \sin b + 7 \sin c - 9 \sin d = 0.
$$
If $\cos(b+c) \neq 0$, determine the value of $\frac{\cos(a+d)}{\cos(b+c)}$.
$$
2 \cos a + 6 \cos b + 7 \cos c + 9 \cos d = 0, \\
2 \sin a - 6 \sin b + 7 \sin c - 9 \sin d = 0.
$$
If $\cos(b+c) \neq 0$, determine the value of $\frac{\cos(a+d)}{\cos(b+c)}$.