It is proved that: (sin20 ° - √ (1-sin178; 20 °) / √ (1-2sin20 ° cos20 °) = - 1

It is proved that: (sin20 ° - √ (1-sin178; 20 °) / √ (1-2sin20 ° cos20 °) = - 1

[sin20°-√(1-sin²20°)]/√(1-2sin20°cos20°)=[sin20°-√(cos²20°)]/√(sin²20°+cos²20°-2sin20°cos20°)=(sin20°-|cos20°|)/√[(sin20°-cos20°)²]=(sin20°-cos20°)/|sin20...

（√1+2sin20°cos20°）/（sin20°+√1-sin^2 20°）
Simplification

[√(1+2sin20cos20）]/[sin20+√(1-sin^2 20)]=[√(sin^2 20+cos^2 20+2sin20cos20)]/[sin20+√(cos^2 20)]={√[(sin20+cos20)^2]}/[sin20+cos20]={sin20+cos20}/[sin20+cos20]=1

sin²1°+sin²2°+sin²3°+.+sin²89°= -1
Why?

sin²1°+sin²2°+sin²3°+.+sin²89°
=sin²1°+cos²1°+sin²2°+cos²2°+----+sin²45°
=44+1/2
=89/2