Автор - DrewPackIe

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

Ответ

Проверено экспертом

Автор - NNNLLL54

frac{tga+1}{1-tga}=frac{frac{sina}{cosa}+1}{1-frac{sina}{cosa}}=frac{sina+cosa}{cosa-sina}=frac{sqrt2cdot (frac{1}{sqrt2}, sina+frac{1}{sqrt2}, cosa)}{sqrt2cdot (frac{1}{sqrt2}, cosa-frac{1}{sqrt2}, sina)}=\\\=frac{sinacdot cosfrac{pi}{4} +cosacdot sinfrac{pi}{4}}{cosacdot cosfrac{pi}{4}-sinacdot sinfrac{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}=sinfrac{pi}{4}=cosfrac{pi}{4}

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