$$
\frac{1}{\cos k \cos (k+1)}=\frac{1}{\sin 1}\frac{\sin 1}{\cos k \cos (k+1)}=\frac{1}{\sin 1} \frac{\sin(k+1)\cos(k)-\sin(k)\cos(k+1)}{\cos k \cos (k+1)} \\ =\frac{1}{\sin 1} \left(\tan(k+1)-\tan (k)\right)
$$
$$
\sin 1\; \sum_{k=0}^{88}\frac{1}{\cos k \cos (k+1)}=\tan(89)-\tan(0)=\frac{\cos(1)}{\sin(1)}
$$