Question

In any $\Delta ABC,$ the least value of $\left( {\dfrac{{{{\sin }^2}A + \sin A + 1}}{{\sin A}}} \right)$ is

A. 3

B. $\sqrt 3$

C. $9$

D. None of these

Answer 1

First of all, split the given function and cancel the common terms. Then use algebraic identities to form the function in terms of a square value. As the value of a square is always greater than zero, we can easily find the least value of the given function. So, use this concept to reach the solution of the given problem.

Complete step-by-step answer:
Let the given triangle $\Delta ABC$ as shown in the below figure:

Let $F\left( A \right) = \left( {\dfrac{{{{\sin }^2}A + \sin A + 1}}{{\sin A}}} \right)$

Separating the fractions, we have

$\Rightarrow F\left( A \right) = \dfrac{{{{\sin }^2}A}}{{\sin A}} + \dfrac{{\sin A}}{{\sin A}} + \dfrac{1}{{\sin A}}$

Cancelling the common terms, we have

$\Rightarrow F\left( A \right) = \sin A + 1 + \dfrac{1}{{\sin A}}$

$\Rightarrow F\left( A \right) = \sin A + \dfrac{1}{{\sin A}} + 1$

We know that $a + \dfrac{1}{a} = {\left( {\sqrt a - \dfrac{1}{{\sqrt a }}} \right)^2} + 2$, by using this formula we get

$\Rightarrow F\left( A \right) = {\left( {\sqrt {\sin A} - \dfrac{1}{{\sqrt {\sin A} }}} \right)^2} + 2 + 1$

$\Rightarrow F\left( A \right) = {\left( {\sqrt {\sin A} - \dfrac{1}{{\sqrt {\sin A} }}} \right)^2} + 3..................................\left( 1 \right)$

We know that, the value of ${\left( {\sqrt {\sin A} - \dfrac{1}{{\sqrt {\sin A} }}} \right)^2}$ is always greater or equal to zero i.e., ${\left( {\sqrt {\sin A} - \dfrac{1}{{\sqrt {\sin A} }}} \right)^2} \geqslant 0$.

Adding ‘3’ on both sides in ${\left( {\sqrt {\sin A} - \dfrac{1}{{\sqrt {\sin A} }}} \right)^2} \geqslant 0$, we have

$\Rightarrow {\left( {\sqrt {\sin A} - \dfrac{1}{{\sqrt {\sin A} }}} \right)^2} + 3 \geqslant 0 + 3$

$\Rightarrow {\left( {\sqrt {\sin A} - \dfrac{1}{{\sqrt {\sin A} }}} \right)^2} + 3 \geqslant 3..................................\left( 2 \right)$

By using equations (1) and (2), we have

$\Rightarrow F\left( A \right) = {\left( {\sqrt {\sin A} - \dfrac{1}{{\sqrt {\sin A} }}} \right)^2} + 3 \geqslant 3$

$\therefore F\left( A \right) = \left( {\dfrac{{{{\sin }^2}A + \sin A + 1}}{{\sin A}}} \right) \geqslant 3$

Therefore, clearly the least value of $\left( {\dfrac{{{{\sin }^2}A + \sin A + 1}}{{\sin A}}} \right)$ is 3.

So, option (A) is correct.

Note: In case we are unable to form the given function in terms of a square value, then use the second derivative test for the function to find the greatest and least value of the given function.