Question

It is given that ${{\sin }^{2}}\theta =\dfrac{4xy}{{{\left( x+y \right)}^{2}}}$ , this expression will be true if and only if :
A. $x-y\ne 0$
B. $x=-y$
C. $x+y\ne 0$
D. $x\ne 0,y\ne 0$

Answer 1

Hint: We will first find the range of ${{\sin }^{2}}\theta $ using the first Pythagorean identity that is ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$ and then we will put the value from the question of ${{\sin }^{2}}\theta $ and then we will solve the inequality. After that, we will see how the fraction will be defined when the denominator should not be 0, and finally we will check from the options that satisfy these conditions and get the answer.

Complete step by step answer:

First of all, let’s find the range of the function: ${{\sin }^{2}}\theta $ , we will start with the basic Pythagorean Identity:
$\begin{align}
  & {{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1\Rightarrow {{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta \\
 & \cos \theta =\sqrt{1-{{\sin }^{2}}\theta } \\
\end{align}$
As we know that the value in the square root is always greater than or equal to $0$ .
Therefore: $1-{{\sin }^{2}}\theta \ge 0\Rightarrow 1\ge {{\sin }^{2}}\theta $ which means that $\sin \theta \in \left[ -1,1 \right]$
So, we have : ${{\sin }^{2}}\theta \ge 1$ , we will put the value of ${{\sin }^{2}}\theta $ in the inequality from the question, as we are given in the question that ${{\sin }^{2}}\theta =\dfrac{4xy}{{{\left( x+y \right)}^{2}}}$ ,
Now: $\dfrac{4xy}{{{\left( x+y \right)}^{2}}}\ge 1$ ,
Now, we will multiply both the sides with ${{\left( x+y \right)}^{2}}$ and we then will get:$4xy\ge {{\left( x+y \right)}^{2}}$
We will now apply the following property : ${{\left( a+b \right)}^{2}}={{a}^{2}}+{{b}^{2}}+2ab$ therefore,
$4xy\ge {{\left( x+y \right)}^{2}}\Rightarrow 4xy\ge {{x}^{2}}+{{y}^{2}}+2xy$ ,
We will now subtract $4xy$ from both the sides of inequality and therefore: $\begin{align}
  & \Rightarrow 4xy\ge {{x}^{2}}+{{y}^{2}}+2xy\Rightarrow 0\ge {{x}^{2}}+{{y}^{2}}+2xy-4xy \\
 & \Rightarrow {{x}^{2}}+{{y}^{2}}-2xy\ge 0 \\
\end{align}$
We will now apply the following property : ${{a}^{2}}+{{b}^{2}}-2ab={{\left( a-b \right)}^{2}}$ therefore,
${{\left( x-y \right)}^{2}}\ge 0$ that means $\left( x-y \right)\ge 0\text{ }..........\text{ Equation 1}\text{.}$
We will check for option A it is given that $x-y\ne 0$ but it will be rejected as it is given that the equation 1 that $x-y$ maybe equal to $0$.
Now we know that in $\dfrac{4xy}{{{\left( x+y \right)}^{2}}}$ , for fraction to be defined the denominator should not be equal to $0$ .
Now from option B , we have $x=-y$ , this is same as $x+y=0$ which is not possible as seen in the above statement.
As we saw that for $\dfrac{4xy}{{{\left( x+y \right)}^{2}}}$ to be defined $\left( x+y \right)$ must not be equal to $0$ .

Hence, the option C is correct that is $x+y\ne 0$.

Note:
Remember that in ${{n}^{2}}$ , $n$ can be either positive or negative but in case of $\sqrt{n}$ , $n$ will be always positive. We need to keep this in our mind in order to find the inequality as it tells us when is a function defined and when it is not.