How do you determine if #secx*tanx# is an even or odd function?

Question

Ariana Alvarez · Accepted Answer

#sec(x)*tan(x)# is an odd function.

An even function is one for which #f(-x) = f(x)# for all #x# in its domain. An odd function is one for which #f(-x) = -f(x)# for all #x# in its domain. Let's begin at: #cos(-x) = cos(x)# #sin(-x) = -sin(x)# #sec(x) = 1/cos(x)# #tan(x) = sin(x)/cos(x)# Next up is: #sec(-x)*tan(-x)# #= 1/cos(-x)*sin(-x)/cos(-x)# #= 1/cos(x)*(-sin(x))/cos(x)# #= -(1/cos(x)*sin(x)/cos(x))# #= -sec(x)*tan(x)# So #sec(x)*tan(x)# is an odd function.

Xavier Alonso · Answer

undefined To determine if $ \sec(x) \tan(x) $ is an even or odd function, we can analyze its behavior under the transformations $ x \to -x $.

For an even function:
$$ f(-x) = f(x) $$

For an odd function:
$$ f(-x) = -f(x) $$

Let's evaluate $ \sec(x) \tan(x) $ under the transformation $ x \to -x $:

$$ \sec(-x) \tan(-x) $$

Using the even/odd properties of secant and tangent functions:
$$ \sec(-x) = \sec(x) $$
$$ \tan(-x) = -\tan(x) $$

Substituting these into $ \sec(-x) \tan(-x) $:
$$ \sec(-x) \tan(-x) = \sec(x) (-\tan(x)) = -\sec(x) \tan(x) $$

Since $ \sec(-x) \tan(-x) = -\sec(x) \tan(x) $, the function $ \sec(x) \tan(x) $ is an odd function.