求解决一个三角恒等变形问题，请给出化解过程。【注意：下面的等式用的是角度制】

(sin³80°-sin70°sin60°)/(sin²80°cos80°)=sin80°/cos80°-sin70°sin60°/(sin²80°cos80°)=cos10°/sin10°-(cos10°-cos130°)/(2sin²80°cos80°)=cos10°/sin10°-(cos10°+cos50°)/(sin80°sin160°)=cos10°/sin10°-(cos10°+sin40°)/(cos10°sin20°)=cos10°/sin10°-1/sin20°-2cos20°/cos10°=2cos²10°/sin20°-1/sin20°-2cos20°/cos10°=cos20°/sin20°-2cos20°/cos10°=2cos²20°/sin40°-2cos20°/sin80°=2cos²20°/sin40°-cos20°/(sin40°cos40°)=(2cos²20°cos40°-cos20°)/(sin40°cos40°)=[(cos40°+1)cos40°-cos20°]/(sin40°cos40°)=(cos²40°+cos40°-cos20°)/(sin40°cos40°)=(cos²40°-2sin30°sin10°)/(sin40°cos40°)=(cos²40°-cos80°)/(sin40°cos40°)=[cos²40°-(cos²40°-sin²40°)]/(sin40°cos40°)=sin²40°/(sin40°cos40°)=tan40°∴sin²80°cos80°/(sin³80°-sin70°sin60°)=1/tan40°=tan50°过程应该能够简化