$5.26 \operatorname{ctg} x-\operatorname{tg} x=2\left(\frac{1}{\operatorname{tg} x+1}+\frac{1}{\operatorname{tg} x-1}\right)=4$ :

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$5.26 \cot x - \tan x = 2\left(\frac{1}{\tan x + 1} + \frac{1}{\tan x - 1}\right) = 4$ :
5.26 We have

$$
\begin{aligned}
& \frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}+2 \frac{\tan x-1+\tan x+1}{\tan^{2} x-1}=4 \\
& \sin x \neq 0, \cos x \neq 0,|\tan x| \neq 1 \\
& \frac{\cos^{2} x-\sin^{2} x}{\sin x \cos x}+\frac{4 \tan x}{\tan^{2} x-1}=4 ; \frac{2 \cos 2 x}{\sin 2 x}-2 \tan 2 x=4 \\
& \cot 2 x-\tan 2 x=2 ; \frac{\cos 2 x}{\sin 2 x}-\frac{\sin 2 x}{\cos 2 x}=2 ; \frac{2 \cos 4 x}{\sin 4 x}=2
\end{aligned}
$$

$$
\cot 4 x=1 ; 4 x=\frac{\pi}{4}+\pi k ; x=\frac{\pi}{16}(4 k+1) .
$$

Answer: $x=\frac{\pi}{16}(4 k+1), k \in \mathbb{Z}$.