Question

How do you determine if $\sec x.\tan x$ is an even or odd function?

Answer 1

A function $f\left( x \right)$ is said to be even, if it satisfies $f\left( x \right)=f\left( -x \right).$ A function is said to be odd, if it satisfies $f\left( -x \right)=-f\left( x \right).$ We use these conditions to check if the given trigonometric function is even or odd.

Complete step by step answer:
Let us take the given function $\sec x.\tan x$ into consideration.
Suppose that $f\left( x \right)=\sec x.\tan x$
Now we have to check if this function $f\left( x \right)$ is an even or odd function.
For that we use the following conditions:
A function $f\left( x \right)$ is said to be an even function if it satisfies the requirement, which is given as $f\left( -x \right)=f\left( x \right).$
A function $f\left( x \right)$ is said to be an odd function if it satisfies the requirement, which is given as $f\left( -x \right)=-f\left( x \right).$
So, here, we have to check if the function $f\left( x \right)=\sec x.\tan x$ satisfies either of the above conditions.
First, let us verify the condition for the function.
That is, if $f\left( -x \right)=f\left( x \right)$ is satisfied.
Now,
$\Rightarrow f\left( x \right)=\sec x.\tan x$
We know that $\sec x=\dfrac{1}{\cos x}$ and $\tan x=\dfrac{\sin x}{\cos x}.$
Also, we have learnt that $\cos x$ is even and $\sin x$ is odd.
That is, $\cos \left( -x \right)=\cos x$ and $\sin \left( -x \right)=-\sin x.$
Now we take,
$\Rightarrow f\left( -x \right)=\sec \left( -x \right)\tan \left( -x \right)$
We are using the above written facts,
$\Rightarrow f\left( -x \right)=\dfrac{1}{\cos \left( -x \right)}\dfrac{\sin \left( -x \right)}{\cos \left( -x \right)}.$
Using the above identities will give us,
$\Rightarrow f\left( -x \right)=\dfrac{1}{\cos x}\dfrac{-\sin x}{\cos x}.$
We are allowed to write this as,
$\Rightarrow f\left( -x \right)=-\dfrac{1}{\cos x}\dfrac{\sin x}{\cos x}.$
That is,
$\Rightarrow f\left( -x \right)=-\sec x.\tan x.$
We know that $-f\left( x \right)=-\sec x\tan x.$
We see that $-f\left( x \right)=f\left( -x \right).$
That is, the function satisfies the condition for an even function.
So, we do not have to check the condition for an odd function.
Hence, the given function $\sec x\tan x$ is an even function.

Note:
Since $\sec x=\dfrac{1}{\cos x},$ $\tan x=\dfrac{\sin x}{\cos x},$ $\cos \left( -x \right)=\cos x$ and $\sin \left( -x \right)=-\sin x,$ we are led to an important fact that $\sec \left( -x \right)=\dfrac{1}{\cos \left( -x \right)}=\dfrac{1}{\cos x}=\sec x.$ Also, $\tan \left( -x \right)=\dfrac{\sin \left( -x \right)}{\cos \left( -x \right)}=\dfrac{-\sin x}{\cos x}=-\tan x.$
That is, from above, we can see that $\sec x$ is an even function and $\tan x$ is an odd function.
Therefore, it is clear that the given function $f\left( x \right)=\sec x\tan x$ is a product of an even and an odd function.
Hence, it is proved that the given function, being a product of an even and odd functions, is an even function.