Тригонометрия arctg(ctg4)

Решите задачу:

$arcctg(ctga)=a\; ,\; \; esli\; \; 0\ \textless \ a\ \textless \ \pi \; ,\; to\; est'\; \; 0^\circ\ \textless \ a\ \textless \ 180^\circ\\\\\\4\; radiana= \frac{4\cdot 180^\circ }{\pi }= \frac{720^\circ }{3,14}\approx 229^\circ =180^\circ +49^\circ \ \textgreater \ 180^\circ \\\\ctg4=ctg(\pi+(4-\pi ))=ctg(4-\pi )\; \; ,\; \; 0\ \textless \ 4-\pi \ \textless \ \pi \\\\arcctg(ctg4)=arcctg(ctg(4-\pi ))=4-\pi \\$

$arctga= \frac{\pi }{2}-arcctga\; \; \to \\\\arctg(ctg4)= \frac{\pi }{2}-arcctg(ctg4)=\frac{\pi }{2}-(4-\pi )=\frac{3\pi }{2}-4$