====== 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. 

=====Question 4(v)=====
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.

=====Question 4(vi)=====
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.

=====Question 4(vii)=====
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.

=====Question 4(viii)=====
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.

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