Question

tg25градусов+tg20градусов/1-tg25градусов tg20градусов

Answer 1

Объяснение:

$\frac{tg25^0+tg20^0}{1-tg25^0*tg20^0}=\frac{\frac{sin25^0}{cos25^0}+\frac{sin20^0}{cos20^0} }{1-\frac{sin25^0*sin20^0}{cos25^0*cos20^0} } =\frac{\frac{sin25^0*cos20^0+sin20^0*cos25^0}{cos25^0*cos20^0} }{\frac{cos25^0*cos20^0-sin25^0*sin20^0}{cjs25^0*cos20^0} } =\frac{sin(25^0+20^0)}{cos(25^0+20^0)} =\\=\frac{sin45^0}{cos45^0}=tg45^0=1.$