题目内容
(1+tan21°)(1+tan22°)(1+tan23°)(1+tan24°)的值是
______.
根据tan45°=tan(21°+24°)=
=1
得到tan21°+tan24°=1-tan21°tan24°①;同理得到tan22°+tan23°=1-tan22°tan23°②;
则原式=[(1+tan21°)(1+tan24°)][(1+tan22°)(1+tan23°)]
=(1+tan24°+tan21°+tan24°tan21°)(1+tan22°+tan23°+tan22°tan23°)
=(1+1-tan24°tan21°+tan24°tan21°)(1+1-tan22°tan23°+tan22°tan23°)=4
故答案为:4.
| tan21°+tan24° |
| 1-tan21°tan24° |
得到tan21°+tan24°=1-tan21°tan24°①;同理得到tan22°+tan23°=1-tan22°tan23°②;
则原式=[(1+tan21°)(1+tan24°)][(1+tan22°)(1+tan23°)]
=(1+tan24°+tan21°+tan24°tan21°)(1+tan22°+tan23°+tan22°tan23°)
=(1+1-tan24°tan21°+tan24°tan21°)(1+1-tan22°tan23°+tan22°tan23°)=4
故答案为:4.
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