证明恒等式arctanx+arccotx=π/2 , f(x) = arctanx+arccotx, 则有f'(x) = 1/(1 + x^2) - 1/(1 + x^2) = 0,f(x) = arctanx+arccotx,则有f'(x) = 1/(1 + x^2) - 1/(1 + x^2) = 0,所以由那个定理,f(x)是常数.把x = 1代入,得到f(1) = arctan 1 + ar
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![证明恒等式arctanx+arccotx=π/2 , f(x) = arctanx+arccotx, 则有f'(x) = 1/(1 + x^2) - 1/(1 + x^2) = 0,f(x) = arctanx+arccotx,则有f'(x) = 1/(1 + x^2) - 1/(1 + x^2) = 0,所以由那个定理,f(x)是常数.把x = 1代入,得到f(1) = arctan 1 + ar](/uploads/image/z/1697593-49-3.jpg?t=%E8%AF%81%E6%98%8E%E6%81%92%E7%AD%89%E5%BC%8Farctanx%2Barccotx%3D%CF%80%2F2+%2C+f%28x%29+%3D+arctanx%2Barccotx%2C+%E5%88%99%E6%9C%89f%27%28x%29+%3D+1%2F%281+%2B+x%5E2%29+-+1%2F%281+%2B+x%5E2%29+%3D+0%2Cf%28x%29+%3D+arctanx%2Barccotx%2C%E5%88%99%E6%9C%89f%27%28x%29+%3D+1%2F%281+%2B+x%5E2%29+-+1%2F%281+%2B+x%5E2%29+%3D+0%2C%E6%89%80%E4%BB%A5%E7%94%B1%E9%82%A3%E4%B8%AA%E5%AE%9A%E7%90%86%2Cf%28x%29%E6%98%AF%E5%B8%B8%E6%95%B0.%E6%8A%8Ax+%3D+1%E4%BB%A3%E5%85%A5%2C%E5%BE%97%E5%88%B0f%281%29+%3D+arctan+1+%2B+ar)
证明恒等式arctanx+arccotx=π/2 , f(x) = arctanx+arccotx, 则有f'(x) = 1/(1 + x^2) - 1/(1 + x^2) = 0,f(x) = arctanx+arccotx,则有f'(x) = 1/(1 + x^2) - 1/(1 + x^2) = 0,所以由那个定理,f(x)是常数.把x = 1代入,得到f(1) = arctan 1 + ar
证明恒等式arctanx+arccotx=π/2 , f(x) = arctanx+arccotx, 则有f'(x) = 1/(1 + x^2) - 1/(1 + x^2) = 0,
f(x) = arctanx+arccotx,
则有f'(x) = 1/(1 + x^2) - 1/(1 + x^2) = 0,
所以由那个定理,f(x)是常数.把x = 1代入,得到
f(1) = arctan 1 + arccot 1 = π/2
所以f(x) = arctanx + arccotx = π/2
这个题的答案为什么要令f(1)=π/2?可以随便假设一个常数?
证明恒等式arctanx+arccotx=π/2 , f(x) = arctanx+arccotx, 则有f'(x) = 1/(1 + x^2) - 1/(1 + x^2) = 0,f(x) = arctanx+arccotx,则有f'(x) = 1/(1 + x^2) - 1/(1 + x^2) = 0,所以由那个定理,f(x)是常数.把x = 1代入,得到f(1) = arctan 1 + ar
那个f'(x)就相当于导数,倒数为零就意味着f(x)的图像为一条水平线,即f(x)为一常数,所以无论是谁都得TT/2