Question

решить
(tg(2x-п/6)+1) / (1-tg(2x-п/6))

Answer 1

$\frac{tga+1}{1-tga}=\frac{\frac{sina}{cosa}+1}{1-\frac{sina}{cosa}}=\frac{sina+cosa}{cosa-sina}=\frac{\sqrt2\cdot (\frac{1}{\sqrt2}\, sina+\frac{1}{\sqrt2}\, cosa)}{\sqrt2\cdot (\frac{1}{\sqrt2}\, cosa-\frac{1}{\sqrt2}\, sina)}=\\\\\\=\frac{sina\cdot cos\frac{\pi}{4} +cosa\cdot sin\frac{\pi}{4}}{cosa\cdot cos\frac{\pi}{4}-sina\cdot sin\frac{\pi}{4}}=\frac{sin(a+\frac{\pi}{4})}{cos(a+\frac{\pi}{4})}=tg(a+\frac{\pi}{4})\; ;$

$a=2x-\frac{\pi}{6}\; \; \Rightarrow \; \; \; \frac{tg(2x-\frac{\pi}{6})+1}{1-tg(2x-\frac{\pi}{6})}=tg((2x-\frac{\pi}{6})+\frac{\pi}{4})=tg(2x+\frac{\pi}{12})$

$P.S.\qquad \frac{1}{\sqrt2}=\frac{\sqrt2}{2}=sin\frac{\pi}{4}=cos\frac{\pi}{4}$