Đáp án:
$1$
Giải thích các bước giải:
`\qquad tan10°.tan25°+tan25°.tan55°+tan55°.tan10°`
`=tan10°. (tan55°+tan25°)+tan55°.tan25°`
`=tan10°. ({sin55°}/{cos55°}+{sin25°}/{cos25°})+{sin55°}/{cos55°}.{sin25°}/{cos25°}`
`=cot80°. {sin55°cos25°+cos55°sin25°}/{cos55°cos25°}+{1/ 2 . [cos(55°-25°)-cos(55°+25°)]}/{cos55°cos25°}`
`={cos80°}/{sin80°} . {sin80°}/{cos55°.cos25°}+1/ 2 . {cos30°-cos80°}/{cos55°.cos25°}`
`={cos80°}/{cos55°.cos25°}+1/ 2 . {cos30°-cos80°}/{cos55°.cos25°}`
`={2cos80°+cos30°-cos80°}/{2cos55°.cos25°}`
`={cos80°+cos30°}/{2. 1/ 2 .[cos(55°+25°)+cos(55°-25°)]}`
`={cos80°+cos30°}/{cos80°+cos30°}=1`
Vậy `tan10°.tan25°+tan25°.tan55°+tan55°.tan10°=1`
