====== Question 2, Exercise 9.1 ======
Solutions of Question 2 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 2(i)=====
Find the maximum and minimum values of the reciprocal trigonometric function: $y=\dfrac{1}{4+3 \operatorname{Sin} \theta}$

** Solution. **

We know
\begin{align*} -1 \leq \operatorname{Sin} \theta \leq 1 \end{align*}
Multiplying with $3$
\begin{align*}  -3 \leq 3 \operatorname{Sin} \theta \leq 3 \end{align*}
Adding $4$
\begin{align*}
 & 1 \leq 4+3 \operatorname{Sin} \theta \leq 7 \\
\implies & 1 \geq \frac{1}{4+3 \operatorname{Sin}} \theta \geq \frac{1}{7} \\
\implies & \frac{1}{7} \leq \frac{1}{4+3 \operatorname{Sin}} \theta\leq 1 \\
\end{align*}
Hence maximum value $(M) = 1$ \\
and minimum value $(m) = \dfrac{1}{7}$. GOOD
=====Question 2(ii)=====
Find the maximum and minimum values of the reciprocal trigonometric function: $y=\dfrac{1}{\frac{1}{2}-5 \operatorname{Cos} \theta}$

** Solution. **

{{ :math-11-nbf:sol:unit09:math-11-nbf-ex9-1-q2_ii_.png?400 |Graph of y}}

**From the graph, we see that given $y$ is not bounded and hence its maximum and minimum value doesn't exist.** 

As $y=\dfrac{1}{\frac{1}{2}-5 \operatorname{Cos} \theta}$ don't have any maximum or minimum value so asking to find out the maximum and minimum values is not appropriate.

=====Question 2(iii)=====
Find the maximum and minimum values of the reciprocal trigonometric function: $y=\dfrac{1}{\frac{1}{3}-4 \sin (2 \theta-5)}$

** Solution. **

**Same as Question 2(ii), we see that given $\dfrac{1}{\frac{1}{3}-4 \sin (2 \theta-5)}$ is not bounded and hence its maximum and minimum value doesn't exist.**

As $\dfrac{1}{\frac{1}{3}-4 \sin (2 \theta-5)}$ don't have any finite maximum or minimum value so asking to find out the maximum and minimum values is not appropriate.
=====Question 2(iv)=====
Find the maximum and minimum values of the reciprocal trigonometric function: $y=\dfrac{1}{3+\frac{2}{5} \sin (5 \theta-7)}$

** Solution. **
	
We know
\begin{align*} -1 \leq \operatorname{Sin(5 \theta-7)} \theta \leq 1 \end{align*}
Multiplying by \(\frac{2}{5}\):  \\
\begin{align*}  &-\frac{2}{5} \leq \frac{2}{5} \operatorname{Sin(5 \theta-7)} \theta \leq \frac{2}{5}\end{align*}
Adding 3:\\  
\begin{align*}
 & 3 - \frac{2}{5} \leq 3 + \frac{2}{5} \operatorname{Sin(5 \theta-7)} \theta \leq 3 + \frac{2}{5}\\
\implies & \dfrac{13}{5} \geq \frac{1}{4+3 \operatorname{Sin(5 \theta-7)}} \theta \geq \dfrac{17}{5} \\
\implies & \dfrac{5}{13} \geq \frac{1}{4+3 \operatorname{Sin(5 \theta-7)}} \theta \geq \dfrac{5}{17} \\
\end{align*}
Hence maximum value $(M) = \dfrac{5}{13}$ \\
and minimum value $(m) = \dfrac{5}{17}$.
====Go to ====

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