Trigonometric treat

Evaluate the sum:

$$\frac{1}{\cos 0^\circ \cos 1^\circ}+\frac{1}{\cos 1^\circ \cos 2^\circ}+ \frac{1}{\cos 2^\circ \cos 3^\circ}+\cdots+ \frac{1}{\cos 88^\circ \cos 89^\circ}$$

Solution

We successively have:

$$\begin{aligned}
\frac{1}{\cos k \cos (k+1)} &=\frac{1}{\sin  1}\frac{\sin 1}{\cos k \cos (k+1)} \\
 &= \frac{1}{\sin 1}\frac{\sin (k+1)\cos k - \sin k \cos (k+1)}{\cos k \cos (k+1)}\\
 &= \frac{1}{\sin 1}\left [ \tan (k+1)- \tan k  \right ]
\end{aligned}$$

Summing telescopically we have that $$\sum_{k=0}^{88}\frac{1}{\cos k \cos (k+1)}=\frac{1}{\sin 1} [\tan 89-\tan 0]=\frac{\cos 1}{\sin^2 1}$$ since $\tan 89 =\cot 1$.

The exercise can also be found in mathematica.gr