Question

2tg²π/6 -ctg²π/3 -sinπ/4×cosπ/3 найдите значение выражения

Answer 1

$2tg^{2}\frac{\pi }{6}-Ctg^{2}\frac{\pi }{3} -Sin\frac{\pi }{4}*Cos\frac{\pi }{3}=2*(\frac{1}{\sqrt{3}})^{2} -(\frac{1}{\sqrt{3}})^{2}-\frac{\sqrt{2}}{2}*\frac{1}{2}=\\\\=2*\frac{1}{3}-\frac{1}{3} -\frac{\sqrt{2}}{4}=\frac{1}{3}-\frac{\sqrt{2}}{4}=\frac{4-3\sqrt{2}}{12}$