Tg(5n/4+x)+tg(5n/4-x)​

Решите задачу:

$tg(\frac{5\pi }{4}+x)+tg(\frac{5\pi }{4}-x)=\frac{sin\frac{5\pi }{2}}{cos(\frac{5\pi }{4}+x)\cdot cos(\frac{5\pi }{4}-x)}=\\\\=\frac{1}{\frac{1}{2}\cdot (cos\frac{5\pi }{2}+cos2x)}}=\frac{2}{1+cos2x}=\frac{2}{2cos^2x}=\frac{1}{cos^2x}=1+tg^2x\\\\\\\\P.S.\; \; \; \; tg(\alpha +\beta )=\frac{sin(\alpha +\beta )}{cos\alpha \cdot cos\beta }\; \; \; ,\; \; \; cos^2\alpha =\frac{1+cos2\alpha }{2}\; \; ,\\\\1+tg^2\alpha =\frac{1}{cos^2\alpha }$

Tg(5n/4+x)+tg(5n/4-x)​

Tg(5n/4+x)+tg(5n/4-x)