设f(x) = x - sinx,g(x) = tanx - x,x∈(0,π/2)
f'(x) = 1 - cosx > 0
g'(x) = sec²x - 1 > 0
由于f(x)和g(x)在(0,π/2)上都是单调递增函数
所以f(x) > f(0) = 0,g(x) > g(0) = 0
==> x - sinx > 0 ==> x > sinx
==> tanx - x > 0 ==> tanx > x
∴sinx < x < tanx,x∈(0,π/2)sec²是什么啊secx = 1/cosx，secx是正割函数，tanx的导数就是sec²x sec²x = 1/cos²x