Question

ctg240°+tg300°-sin(-225°)-cos495°​

Answer 1

Ответ:

Объяснение:

ctg240°+tg300°-sin(-225°)-cos495°​=

=сtg(180°+60°)+tg(360°-300°)+sin(225°)-cos(360°+135°)=

=ctg60°+(-tg60°)+sin(180+45°)-cos(135°)=

ctg60°-tg60°-sin45°-cos(180°-45°)=√3/3-√3-√2/2-(-cos45°)=

=-2√3/3-√2/2+√2/2=-2√3/3.