Найдите tgα, если \frac{3sin\alpha–5cos\alpha+2}{sin\alpha+3cos\alpha+6}=\frac{1}{3}.
Источник: mathege
Решение:
\frac{3sin\alpha–5cos\alpha+2}{sin\alpha+3cos\alpha+6}=\frac{1}{3}\\(3sin\alpha–5cos\alpha+2)\cdot 3=(sin\alpha+3cos\alpha+6)\cdot 1\\9sin\alpha–15cos\alpha+6=sin\alpha+3cos\alpha+6\\9sin\alpha-sin\alpha=3cos\alpha+15cos\alpha+6-6\\8sin\alpha=18cos\alpha\:{\color{Blue} |: 2}\\4sin\alpha=9cos\alpha\:{\color{Blue} |: 4cos\alpha}\\\frac{4sin\alpha}{4cos\alpha}=\frac{9cos\alpha}{4cos\alpha}\\tg\alpha=\frac{9}{4}=2\frac{1}{4}=2,25
Ответ: 2,25.