13. Solve the equation $\frac{1}{\cos x \cos 2 x}+\frac{1}{\cos 2 x \cos 3 x}+\cdots+\frac{1}{\cos 100 x \cos 101 x}=0$.
13. $\tan (k+1) x-\tan k x=\frac{\sin (k+1) x \cdot \cos k x-\cos (k+1) x \sin k x}{\cos (k+1) x \cos k x}=\frac{\sin x}{\cos k x \cos (k+1) x}$.

Then the equation can be transformed into $\frac{1}{\sin x}[(\tan 2 x-\tan x)+(\tan 3 x-\tan 2 x)+\cdots+(\tan 101 x-$ $\tan 101 x)]=0$, which is $\frac{\tan 101 x-\tan x}{\sin x}=0 \Leftrightarrow \frac{\sin 100 x}{\sin x \cos 101 x \cos x}=0, \sin 100 x=0, x=\frac{n \pi}{100}, n \in$ Z. Therefore, the solution to the original equation is $\left\{x \left\lvert\, x=\frac{n \pi}{100}\right., \sin \frac{n \pi}{100} \neq 0, \cos \frac{n \pi}{100} \neq 0, n \in \mathbf{Z}\right\}=\left\{x \left\lvert\, x=\frac{t}{25} \pi\right., t \in\right.$ $\mathbf{Z}$, and $t$ is not a multiple of 25 $\}$.