Question

Let $f\left(x\right)=\text{sec}x+\tan x,g\left(x\right)=\frac{\left[\tan x\right]}{1-\text{sec}x}$.

Statement 1: $g$ is an odd function.

Statement 2: $f$ is neither an odd function nor an even function.

• A. Statement 1 is true
• B. Statement 2 is true
• C. Statements 1 and 2 both are true
• D. Statements 1 and 2 both are false

Answer 1

The correct option is C

Statements 1 and 2 both are true

Explanation for the correct answer:

Checking the truth of the two statements:

Step 1: Given information:

Given that: $f\left(x\right)=\text{sec}x+\tan x,g\left(x\right)=\frac{\left[\tan x\right]}{1-\text{sec}x}$

An odd function is one where $f(-x)=-f\left(x\right)$

An even function is one where $f(-x)=f\left(x\right)$

Step 2: Checking whether $\mathrm{f}$ is an odd or even function

$$\begin{gathered} f(-x)=\text{sec}(-x)+\tan (-x) \\ =\text{sec}x-\tan x \\ \ne -f\left(x\right) \end{gathered}$$

Therefore, $\mathrm{f}$ is neither an odd function nor an even function.

Step 3: Checking if $\mathrm{g}$ is an odd function:

$\begin{array}{rcl}g(-x)& =& \frac{\tan (-x)}{1-\text{sec}(-x)}\\ & =& -\frac{\tan x}{1-\text{sec}x}\\ & =& -g\left(x\right)\end{array}$

Therefore, $\mathrm{g}$ is an odd function.

So, both statement 1 and statement 2 are true.

Hence, option (C) is the correct answer.