Question

Помогите решить эту задачу (не через Photomath):​

Answer 1

$\displaystyle\bf\\\frac{\dfrac{\sqrt{3} }{3} -tg\dfrac{11\pi }{12} }{1-\dfrac{\sqrt{3} }{3} \cdot tg\dfrac{\pi }{12} } =\frac{tg\dfrac{\pi }{6} -tg\Big(\pi -\dfrac{\pi }{12} \Big)}{1-tg\dfrac{\pi }{6} \cdot tg\dfrac{\pi }{12} } =\frac{tg\dfrac{\pi }{6} +tg\dfrac{\pi }{12} }{1-tg\dfrac{\pi }{6} \cdot tg\dfrac{\pi }{12} } =\\\\\\=tg\Big(\frac{\pi }{6} +\frac{\pi }{12} \Big)=tg\frac{3\pi }{12} =tg\frac{\pi }{4} =1$