作业帮三角函数 (5 20:54:8)计算:(1+2cos260^o.cos350^o)^1/2／[cos10^o-(1-cos^2170^o)^1/2]2.己知tan(兀-乄)=a^2,｜cos(兀-乄)｜=-cos乄,求1/cos(兀+乄)的值
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![三角函数 (5 20:54:8)计算:(1+2cos260^o.cos350^o)^1/2／[cos10^o-(1-cos^2170^o)^1/2]2.己知tan(兀-乄)=a^2,｜cos(兀-乄)｜=-cos乄,求1/cos(兀+乄)的值](/uploads/image/z/3714233-41-3.jpg?t=%E4%B8%89%E8%A7%92%E5%87%BD%E6%95%B0+%285+20%3A54%3A8%29%E8%AE%A1%E7%AE%97%3A%281%2B2cos260%5Eo.cos350%5Eo%29%5E1%2F2%EF%BC%8F%5Bcos10%5Eo-%281-cos%5E2170%5Eo%29%5E1%2F2%5D2.%E5%B7%B1%E7%9F%A5tan%28%E5%85%80-%E4%B9%84%29%3Da%5E2%2C%EF%BD%9Ccos%28%E5%85%80-%E4%B9%84%29%EF%BD%9C%3D-cos%E4%B9%84%2C%E6%B1%821%2Fcos%28%E5%85%80%2B%E4%B9%84%29%E7%9A%84%E5%80%BC)
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三角函数 (5 20:54:8)计算:(1+2cos260^o.cos350^o)^1/2／[cos10^o-(1-cos^2170^o)^1/2]2.己知tan(兀-乄)=a^2,｜cos(兀-乄)｜=-cos乄,求1/cos(兀+乄)的值
三角函数 (5 20:54:8)
计算:(1+2cos260^o.cos350^o)^1/2／[cos10^o-(1-cos^2170^o)^1/2]
2.己知tan(兀-乄)=a^2,｜cos(兀-乄)｜=-cos乄,求1/cos(兀+乄)的值
三角函数 (5 20:54:8)计算:(1+2cos260^o.cos350^o)^1/2／[cos10^o-(1-cos^2170^o)^1/2]2.己知tan(兀-乄)=a^2,｜cos(兀-乄)｜=-cos乄,求1/cos(兀+乄)的值
给你第一题吧!第二题的条件可能是错了.你把第二题重新提问一次好了.
