Question

If the value of x is given in the interval as $0 < x \leq \dfrac{\pi }{2} $ then find the minimum value of $ \sin x\text{ + }\csc x $.
A. 0
B. -1
C. 1
D. 2

Answer 1

Hint: According to the question, it has been said that x is an acute angle. So, any trigonometric operation will be naturally a positive quantity. Thus, both sin x and cosec x are positive values. We know, for two positive quantities, Arithmetic Mean of those quantities is greater than or equal to the geometric mean of those quantities, i.e. $\text{A}\text{.M}\text{. }\ge \text{ G}\text{.M}\text{.} $ Apply this inequality to find the minimum value of sinx + cosecx.

Complete step-by-step answer:
It is given that $0 < x \leq \dfrac{\pi }{2} $. Thus, x is an acute angle.
Therefore, both sin x and cosec x are positive quantities. Therefore, we can apply the inequality $\text{A}\text{.M}\text{. }\ge \text{ G}\text{.M}\text{.}$ on the two positive quantities sinx and cosecx, where $0 < x \leq \dfrac{\pi }{2} $.
We also know that cosec x is the reciprocal value of sinx. Thus, we can write,
$\cos \text{ec}x\text{ = }\dfrac{1}{\sin x}$
Applying $\text{A}\text{.M}\text{. }\ge \text{ G}\text{.M}\text{.}$on sinx and cosecx, we get,
$\begin{align}
  & \dfrac{\sin x\text{ + }\cos \text{ec}x}{2}\text{ }\ge \text{ }\sqrt{\sin x\cdot \operatorname{cose}\text{c}x} \\
 & \Rightarrow \text{ }\dfrac{\sin x\text{ + }\cos \text{ec}x}{2}\text{ }\ge \text{ }\sqrt{\sin x\cdot \dfrac{1}{\sin x}}\text{ }\left( \because \text{ }\cos \text{ec}x\text{ = }\dfrac{1}{\sin x}\text{ } \right) \\
 & \Rightarrow \text{ }\dfrac{\sin x\text{ + }\cos \text{ec}x}{2}\text{ }\ge \text{ 1} \\
 & \therefore \text{ }\sin x\text{ + }\cos \text{ec}x\text{ }\ge \text{ 2} \\
\end{align}$
Therefore, the minimum value of sinx + cosecx is 2.
Thus, the correct answer is option A.

Note: Alternatively, this problem can be solved without using the AM-GM inequality. We know that the square value of any quantity is always positive. Thus, we can write ${{\left( \sin x-1 \right)}^{2}}\text{ }\ge \text{ 0}$. Using this inequality, we can subsequently prove,
$\begin{align}
  & {{\left( \sin x-1 \right)}^{2}}\text{ }\ge \text{ 0} \\
 & \Rightarrow \text{ si}{{\text{n}}^{2}}x\text{ }-\text{ 2}\sin x\text{ + 1 }\ge \text{ 0} \\
\end{align}$
As sinx is open at x = 0, so sinx can never be zero. So, dividing by sin x on both sides of the above equation, we get,
$\begin{align}
  & \sin x\text{ }-\text{ 2 + }\dfrac{1}{\sin x}\text{ }\ge \text{ 0} \\
 & \Rightarrow \text{ }\sin x\text{ + }\dfrac{1}{\sin x}\text{ }\ge \text{ 2 } \\
 & \therefore \text{ }\sin x\text{ + }\cos \text{ec}x\text{ }\ge \text{ 2} \\
\end{align}$
Thus, by alternative method as well, we get the minimum value of sinx + cosecx to be equal to 2. Also note that, the range is discontinuous at x = 0, where the value of cosec x is undefined.