4. Zadatak

$$A=\frac{\tg^2x-\ctg^2x}{\tg^3x+\ctg^3x} =\frac{(\tg x-\ctg x)(\tg x+\ctg x)}{(\tg x+\ctg x)(\tg^2x-1+\ctg^2x)} =\frac{\tg x-\ctg x}{\tg^2x+\ctg^2x-1}.\tag{1}$$

Iz $\tg x=4+\ctg x$ izlazi

$$\tg x-\ctg x=4.\tag{2}$$

a odatle kvadriranjem

$$\tg^2x-2+\ctg^2x=16$$ $$\tg^2x+\ctg^2x-1=17.\tag{3}$$

Sada iz (1), (2) i (3) slijedi

$$A=\frac4{17}.$$