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

##### 4 Answers

#### Explanation:

We find:

#tan 1 ~~ 1.5574#

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

So

#### Explanation:

Note that:

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

for all real values of

If

#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

In addition note that

Hence

In particular note that for small

So:

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

including

See below.

#### Explanation:

Note that near

The

#### 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

tangent is an increasing function and is continuous on the interval

Since

So