Question 4(v-viii), Exercise 9.1

Solutions of Question 4(v-viii) of Exercise 9.1 of Unit 09: Trigonometric Functions. This is unit of Model Textbook of Mathematics for Class XI published by National Book Foundation (NBF) as Federal Textbook Board, Islamabad, Pakistan.

Check whether the function is odd or even: $y=\dfrac{\sin ^{2} x}{x+\tan x}$

Solution.

Consider $$y = \frac{\sin^2 x}{x + \tan x}$$ Take $$\begin{align*} y(-x) &= \frac{\big(-\sin x\big)^2}{-x - \tan x} \\ &= \frac{\sin^2 x}{-x - \tan x}\\ & = \frac{\sin^2 x}{-(x + \tan x)}\\ & = -\frac{\sin^2 x}{x + \tan x}\\ &=-y(x)\end{align*}$$ Thus, the given function is odd.

Check whether the function is odd or even: $y=\dfrac{\tan x-\sin x}{\sin^3 x}$

Solution.

Consider $$y = \frac{\tan x - \sin x}{\sin^3 x}$$ Take $$\begin{align*} y(-x) &= \frac{-\tan x - (-\sin x)}{(-\sin x)^3} \\ &= \frac{-\tan x + \sin x}{-\sin^3 x}\\ &= \frac{-(\tan x - \sin x)}{\sin^3 x}\\ &= -\frac{\tan x - \sin x}{\sin^3 x}\\ & = -y(x) \end{align*}$$ Thus, the given function is odd.

Check whether the function is odd or even: $y=\dfrac{\sec x}{x+\tan x}$

Solution.

Consider $$y = \frac{\tan x - \sin x}{\sin^3 x}$$ Take $$\begin{align*} y(-x) &= \frac{-\tan x - (-\sin x)}{(-\sin x)^3} \\ &= \frac{-\tan x + \sin x}{-\sin^3 x}\\ &= \frac{-(\tan x - \sin x)}{\sin^3 x}\\ &=-\frac{\tan x - \sin x}{\sin^3 x}\\ &= -y(x)\end{align*}$$ Thus, the given function is odd.

Check whether the function is odd or even: $y=x^{2} \cdot \sin x -\cot x$

Solution.

Consider $$y = x^2 \cdot \sin x - \cot x$$ Take $$\begin{align*} y(-x) &= (-x)^2 \cdot (-\sin x) - (-\cot x) \\ &= x^2 \cdot (-\sin x) + \cot x \\ &= -x^2 \cdot \sin x + \cot x\\ &=-y(x)\end{align*}$$ Thus, the given function is odd.

< Question 4(i-iv) Question 5(i-v) >