Task B-3.4. If $\operatorname{tg} x=4+\operatorname{ctg} x$, calculate

$$
\frac{\operatorname{tg}^{2} x-\operatorname{ctg}^{2} x}{\operatorname{tg}^{3} x+\operatorname{ctg}^{3} x}
$$
Solution.

$$
A=\frac{\tan^{2} x-\cot^{2} x}{\tan^{3} x+\cot^{3} x}=\frac{(\tan x-\cot x)(\tan x+\cot x)}{(\tan x+\cot x)\left(\tan^{2} x-1+\cot^{2} x\right)}=\frac{\tan x-\cot x}{\tan^{2} x+\cot^{2} x-1}
$$

From $\tan x=4+\cot x$ it follows that

$$
\tan x-\cot x=4
$$

and from squaring this,

$$
\begin{aligned}
& \tan^{2} x-2+\cot^{2} x=16 \\
& \tan^{2} x+\cot^{2} x-1=17
\end{aligned}
$$

(1 point)

Now from (1), (2), and (3) it follows that $A=\frac{4}{17}$.