Recall that cosine is an even function and sine is an odd function. In terms of equations:
$$\cos(-x) = \cos(x)$$
$$\sin(-x) = -\sin(x)$$
We can determine whether each of the other basic trigonometric functions is even, odd, or neither, with just these two facts and the reciprocal identities.
Tangent
Recall that
$$\tan(x) = \frac{\sin(x)}{\cos(x)}$$
We know that sine is odd and cosine is even. Furthermore, by substituting \(x \Rightarrow -x\) in the identities above, we get the also true
$$\cos(x) = \cos(-x)$$
$$\sin(x) = -\sin(-x)$$
We can substitute these into the expression for tangent:
$$\frac{\sin(x)}{\cos(x)} = \frac{-\sin(-x)}{\cos(-x)} = -\tan(-x)$$
Dividing the equation \(\tan(x) = -\tan(-x)\) by \(-1\) gives
$$-\tan(x) = \tan(-x)$$
Thus tangent takes the form \(f(-x) = -f(x)\), so tangent is an odd function.
Cotangent
Taking the reciprocal of the identity shown above gives
$$-\frac{1}{\tan(x)} = \frac{1}{\tan(-x)} \Rightarrow$$
$$-\cot(x) = \cot(-x)$$
Therefore cotangent is also an odd function.
Secant
Start with the even function identity for cosine:
$$\cos(-x) = -\cos(x)$$
Take the reciprocal of both sides of the equation:
$$\frac{1}{\cos(-x)} = -\frac{1}{\cos(x)} \Rightarrow$$
$$\sec(-x) = -\sec(x)$$
Therefore secant is an even function.
At this point, we advise you to ask your students to determine whether cosecant is even or odd with this same procedure as an exercise (note: it is an odd function, and the procedure is basically the same as the one above).