The integer part of $\sum_{k=0}^{44}\frac{1}{\cos(k^\circ)\cos((k+1)^\circ)}$ What is the integer part of the number $$\sum_{k=0}^{44}\frac{1}{\cos (k^\circ)\cos((k+1)^\circ)}$$ 


I tried to solve it using partial fractions but could not get a result. Please help me out.
Notice for any $a &lt; b$, we have

$$\tan b - \tan a = \frac{\sin b\cos a - \sin a\cos b}{\cos b\cos a} = \frac{\sin(b-a)}{\cos b\cos a}$$
This implies
$$\begin{align}\sum_{k=0}^{44} \frac{1}{\cos(k^\circ)\cos((k+1)^\circ)}
&amp;= \frac{1}{\sin 1^\circ}\sum_{k=0}^{44} \left(\tan((k+1)^\circ) - \tan(k^\circ)\right)\\
&amp;= \frac{1}{\sin 1^\circ}\left(\tan(45^\circ) - \tan(0^\circ)\right)\\
&amp;= \frac{1}{\sin\frac{\pi}{180}} \approx 57.29868849855018
\end{align}$$
The integer we seek is $57$.
Playing around.

Let
$s(n)
=\sum_{k=0}^{n}\frac{1}{\cos(k)\cos(k+1)}
$.

Looking at the sum in Wolfy,
it peaks at about 12
and has a period of about 22
with 7 peaks.
The extreme values 
are about $\pm 10$.
This is probably due to
$\pi \approx \frac{22}{7}$.

In particular,
$s(44) \approx 1.9$
so the integer part is 1.

Also,

$\begin{array}\\
\frac{1}{\cos(k)}-\frac1{\cos(k+1)}
&amp;=\frac{\cos(k+1)-\cos(k)}{\cos(k)\cos(k+1)}\\
&amp;=-2\frac{\sin(k+\frac12)\sin(\frac12)}{\cos(k)\cos(k+1)}\\
\end{array}
$

but this doesn't seem to help much.

Can't think of
anything else.