Question

Which is greater, `tan 1` or `tan^(-1) 1` ?

Answer 1

`tan^(-1) 1`

Explanation:

We find:

`tan 1 ~~ 1.5574`

`tan^(-1) 1 = pi/2 ~~ 1.5757`

So `tan^(-1) 1` is slightly greater than `tan 1`

Answer 2

`tan 1 > tan^(-1) 1`

Explanation:

Note that:

`d/(dx) tan x = sec^2 x >= 1`

for all real values of `x` in the domain of `tan x`

If `y = tan^(-1) x`, then:

`tan y = x`

So:

`(dy)/(dx) sec^2 y = 1`

`(dy)/(dx) = 1/(sec^2 y) = 1/(1+tan^2 y) = 1/(1+x^2)`

Hence:

`d/(dx) tan^(-1) x in (0, 1]` for all `x in RR`

So:

`d/(dx) (tan x - tan^(-1) x) = sec^2 x - 1/(1+x^2) >= 0`

for all `x` in the domain of `tan x`

In addition note that `tan x` is defined and continuous in `[0, pi/2)`

Hence `tan x - tan^(-1) x >= 0` for all `x in [0, pi/2)`

In particular note that for small `epsilon > 0` we have `d/(dx) tan x > 1` and `d/(dx) tan^(-1) x < 1`.

So:

`tan x > tan^(-1) x` for all `x in (0, pi/2)`

including `x=1`

Answer 3

See below.

Explanation:

Note that near `x = 0`

`tanx = x + x^3/3 + 2/15 x^5+O(x^7)` and
`tan^-1 x =x - x^3/3+x^5/5+O(x^7)`

The `tanx` series for `x > 0` always sum up and the series for `tan^-1 x` is an alternating series so

`tan gt tan^-1 x` for `0 < x < pi/2`

Answer 4

`tan(1) > tan^-1(1)`

Explanation:

Another approach. This one is pretty much all first semester of trig stuff as long as you've covered graphing tangent.

We know that `tan^-1(1) = pi/4 approx 0.785`.

tangent is an increasing function and is continuous on the interval `-pi/2 < x < pi/2`. Since tangent is increasing on that interval and `pi/4 < 1` we know that `tan(pi/4) < tan(1)`.

Since `tan(pi/4) = 1` we know that `tan(1) > 1`.

So `tan(1) > 1` and `tan^(-1)(1) approx 0.785`, we know that `tan(1) > tan^-1(1)`.