Chapter 0.6, Problem 45E

(a)

To determine

To prove : The quotient of an even function and an odd function is an odd function.

Explanation of Solution

Given information :

The quotient of an even function and an odd function

Formula used :

If a function is even then $f\left(-x\right)=f\left(x\right)$

Proof:

Suppose that we have an even function $f\left(x\right)$

Then, $f\left(-x\right)=f\left(x\right)$

And we have another function $g\left(x\right)$ which is odd

Then, $g\left(-x\right)=-g\left(x\right)$

Now

  $\begin{array}{l}\frac{f\left(-x\right)}{g\left(-x\right)}=\frac{f\left(x\right)}{-g\left(x\right)}\left(\text{Since }f\left(x\right)\text{ is even and }g\left(x\right)\text{is odd}\right)\\ =-\frac{f\left(x\right)}{g\left(x\right)}\end{array}$

Hence we can say that the quotient of an even function and an odd function is an odd function.

(b)

To determine

To prove: $\cot x$ is an odd function of $x$ .

Explanation of Solution

Given information :

  $\cot x$

Formula used :

If a function is odd then $f\left(-x\right)=-f\left(x\right)$

Proof:

Let $f\left(x\right)=\cot x$

Then

  $\begin{array}{l}f\left(-x\right)=\cot \left(-x\right)\\ =\frac{1}{\tan \left(-x\right)}\\ =-\frac{1}{\tan x}\\ =-\cot x\\ =-f\left(x\right)\end{array}$

Hence, $\cot x$ is an odd function of $x$ .