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

Jan 21, 2018

${\tan}^{- 1} 1$

#### Explanation:

We find:

$\tan 1 \approx 1.5574$

${\tan}^{- 1} 1 = \frac{\pi}{2} \approx 1.5757$

So ${\tan}^{- 1} 1$ is slightly greater than $\tan 1$

Jan 25, 2018

$\tan 1 > {\tan}^{- 1} 1$

#### Explanation:

Note that:

$\frac{d}{\mathrm{dx}} \tan x = {\sec}^{2} x \ge 1$

for all real values of $x$ in the domain of $\tan x$

If $y = {\tan}^{- 1} x$, then:

$\tan y = x$

So:

$\frac{\mathrm{dy}}{\mathrm{dx}} {\sec}^{2} y = 1$

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{1}{{\sec}^{2} y} = \frac{1}{1 + {\tan}^{2} y} = \frac{1}{1 + {x}^{2}}$

Hence:

$\frac{d}{\mathrm{dx}} {\tan}^{- 1} x \in \left(0 , 1\right]$ for all $x \in \mathbb{R}$

So:

$\frac{d}{\mathrm{dx}} \left(\tan x - {\tan}^{- 1} x\right) = {\sec}^{2} x - \frac{1}{1 + {x}^{2}} \ge 0$

for all $x$ in the domain of $\tan x$

In addition note that $\tan x$ is defined and continuous in $\left[0 , \frac{\pi}{2}\right)$

Hence $\tan x - {\tan}^{- 1} x \ge 0$ for all $x \in \left[0 , \frac{\pi}{2}\right)$

In particular note that for small $\epsilon > 0$ we have $\frac{d}{\mathrm{dx}} \tan x > 1$ and $\frac{d}{\mathrm{dx}} {\tan}^{- 1} x < 1$.

So:

$\tan x > {\tan}^{- 1} x$ for all $x \in \left(0 , \frac{\pi}{2}\right)$

including $x = 1$

Jan 25, 2018

See below.

#### Explanation:

Note that near $x = 0$

$\tan x = x + {x}^{3} / 3 + \frac{2}{15} {x}^{5} + O \left({x}^{7}\right)$ and
${\tan}^{-} 1 x = x - {x}^{3} / 3 + {x}^{5} / 5 + O \left({x}^{7}\right)$

The $\tan x$ series for $x > 0$ always sum up and the series for ${\tan}^{-} 1 x$ is an alternating series so

$\tan > {\tan}^{-} 1 x$ for $0 < x < \frac{\pi}{2}$

Jan 26, 2018

$\tan \left(1\right) > {\tan}^{-} 1 \left(1\right)$

#### 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 ${\tan}^{-} 1 \left(1\right) = \frac{\pi}{4} \approx 0.785$.

tangent is an increasing function and is continuous on the interval $- \frac{\pi}{2} < x < \frac{\pi}{2}$. Since tangent is increasing on that interval and $\frac{\pi}{4} < 1$ we know that $\tan \left(\frac{\pi}{4}\right) < \tan \left(1\right)$.

Since $\tan \left(\frac{\pi}{4}\right) = 1$ we know that $\tan \left(1\right) > 1$.

So $\tan \left(1\right) > 1$ and ${\tan}^{- 1} \left(1\right) \approx 0.785$, we know that $\tan \left(1\right) > {\tan}^{-} 1 \left(1\right)$.