Question

Assume that u,v E ( -pi/2, pi/2 ). use mean value theorem to prove that |tanu-tanv| >= |u-v|

Answer 1

let f(x)=tan x
f(x) and f'(x) are continuous on E
withot loss of generality let u>v
so there are some values v<w<u such that
\[\frac{\tan u - \tan v}{u-v}=f'(w)=1+\tan^2 w \ge 1\]

Answer 2

@mukushla : Could you explain me a little bit more, please.

Answer 3

@JamesJ @TuringTest

help plz my english is not well :(

Answer 4

@waterineyes

Answer 5

@FoolAroundMath

Answer 6

@JamesJ @Hero could you help me with this

Answer 7

Let f(x) be defined in [a, b ] and differentiable in (a, b). there exists a number c betwween a abd b such that f'(c) = [f(b) - f(a)]/(b-a). That is mean value theorem,

Given f(x) = tan (x) the derivative is f'(x) = sec^2 (x)
Assume u > v. There exist w such v < w < u and
[tan(u) - tan(v)] / (u - v) = sec^2 (x).
since sec (x) either >= 1 or =< -1, sec^2 (x) >= 1
(We can also let sec^2 (x) = 1 + tan^2 (x) >= 1)
So [tan (u) - tan(v)] >= (u - v)

If you don't assume u > v, then we need to use absolute value on both sides.

Answer 8

THank @Mr._To and @mukushla igot it now