Đáp án:

 `1`

Giải thích các bước giải:

 `sin^{2}10^{o}+sin^{2}20^{o}+...+sin170^{o}+sin^{2}180^{o}`

`=(sin^{2}10^{o}+sin^{2}170^{o})+(sin^{2}20^{o}+sin^{2}160^{o})+(sin^{2}30^{o}+sin^{2}150^{o})+(sin^{2}40^{o}+sin^{2}140^{o})+(sin^{2}50^{o}+sin^{2}130^{o})+(sin^{2}60^{o}+sin^{2}120^{o})+(sin^{2}70^{o}+sin^{2}110^{o})+(sin^{2}80^{o}+sin^{2}100^{o})+sin^{2}90^{o}+sin^{2}180^{o}`

Ta có:

`sin^{2}170^{o}=-cos^{2}(180^{o}-170^{o})=-cos^{2}10^{o}`

Tượng tự ta có:

`sin^{2}160^{o}=-cos^{2}20^{o}`

.............

Ta được:

`=(sin^{2}10^{o}-cos^{2}10^{o})+(sin^{2}20^{o}-cos^{2}20^{o})+(sin^{2}30^{o}-cos^{2}30^{o})+(sin^{2}40^{o}-cos^{2}40^{o})+(sin^{2}50^{o}-cos^{2}50^{o})+(sin^{2}60^{o}-cos^{2}60^{o})+(sin^{2}70^{o}-cos70^{o})+(sin^{2}80^{o}-cos^{2}80^{o})+1+0`

`=0+0+0+0+0+0+0+0+1`

`=1`