How do you find the domain & range for #sec theta#?

1 Answer
Mar 5, 2016

Answer:

Domain

#{x| x = (k+1/2) pi, k in RR\\\ZZ}#

Range

#(-oo,1] uu [1,oo)#

Explanation:

Graph of #y = sec(x)#
graph{sec(x) [-20, 20, -10, 10]}
#sec theta = 1/cos theta#. As usual, division of zero is not allowed.

#cos theta = 0# when #theta = pi/2, {3pi}/2, {5pi}/2 ...#

In general, #cos theta = 0# when #theta = (k+1/2) pi#, for #k in ZZ#.

The domain for #sec(theta)# is any real number that

when subtracted #pi/2#, is not an integer multiple of #pi#.

In mathematical notations, it is

#{x| x = (k+1/2) pi, k in RR\\\ZZ}#

Note that the domain of #sec(theta)# and #tan(theta)# are identical.

The since #-1 <= cos(theta) <=1#, you can look at the graph of #y = 1/x#, and close in on the portion of #-1 <= x <=1#.
graph{1/x [-5, 5, -2.5, 2.5]}
You will see that it is either #y >= 1# or #y <= -1#. Similarly, for #sec(theta)#, it is either #sec(theta) >= 1# or #sec(theta) <= -1#.

In mathematical notations, it is

#(-oo,1] uu [1,oo)#