Question

化简:[根号(1-2sin290°cos110°)]/(sin250°+sin20°)

Answer 1

[根号(1-2sin290°cos110°)]/(sin250°+sin20°)

=[根号(1-2sin(180°+110°)cos110°)]/[sin(270°-20°)+sin20°]

=[根号(1+2sin110°cos110°)]/(-cos20°+sin20°)

={根号[(sin110°)^2+(cos110°)^2+2sin110°cos110°]}/(-c...