Question

Consider the statements
Statement1: If $x\cos \theta =y\cos \left( {{120}^{\circ }}+\theta \right)=z\cos \left( {{240}^{\circ }}+\theta \right)$, then $xy+yz+zx=0$
Statement2: Value of $\cos \alpha +\cos \left( {{120}^{\circ }}+\alpha \right)+\cos \left( {{120}^{\circ }}-\alpha \right)=0$
Then which of the above statements is correct?
A. Only 1
B. Only 2
C. Both 1 and 2
D. Neither 1 nor 2

Answer 1

Hint: To solve this problem, we should know the formulae related to $\cos \left( A+B \right)$ and $\cos \left( A-B \right)$. The formulae are given as
$\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$
$\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$
Using these two formulae, we can find if the statement-2 is correct or not. For the statement-1 we should write $x\cos \theta =y\cos \left( {{120}^{\circ }}+\theta \right)=z\cos \left( {{240}^{\circ }}+\theta \right)=k$ and
$x=\dfrac{k}{\cos \theta },y=\dfrac{k}{\cos \left( {{120}^{\circ }}+\theta \right)},z=\dfrac{k}{\cos \left( {{240}^{\circ }}+\theta \right)}$. By considering $xy+yz+zx$ and taking the L.C.M, we get the numerator similar to the statement-2 in which we can use the above formulae to get the answer.

Complete step-by-step answer:
Let us consider the statement-2
$\cos \alpha +\cos \left( {{120}^{\circ }}+\alpha \right)+\cos \left( {{120}^{\circ }}-\alpha \right)$
We know the formulae
$\cos \left( A+B \right)=\cos A\cos B-\sin A\sin B$
$\cos \left( A-B \right)=\cos A\cos B+\sin A\sin B$
Using them in the equation, where $A={{120}^{\circ }}$ and $B=\alpha $, we get
 $\begin{align}
  & \cos \alpha +\cos {{120}^{\circ }}\cos \alpha -\sin {{120}^{\circ }}\sin \alpha +\cos {{120}^{\circ }}\cos \alpha +\sin {{120}^{\circ }}\sin \alpha \\
 & =\cos \alpha +2\cos {{120}^{\circ }}\cos \alpha \\
\end{align}$
We know that $\cos {{120}^{\circ }}=\cos \left( 90+30 \right)=-\sin 30=-\dfrac{1}{2}$
By substituting the value in above equation, we get
$\cos \alpha +2\cos {{120}^{\circ }}\cos \alpha =\cos \alpha -2\times \dfrac{1}{2}\cos \alpha =\cos \alpha -\cos \alpha =0$
Hence, we can write that statement-2 is correct.
Let us consider statement-1.
$x\cos \theta =y\cos \left( {{120}^{\circ }}+\theta \right)=z\cos \left( {{240}^{\circ }}+\theta \right)$.
Let us consider a value k which is equal to the whole equation.
$x\cos \theta =y\cos \left( {{120}^{\circ }}+\theta \right)=z\cos \left( {{240}^{\circ }}+\theta \right)=k$
We can write the individual terms of x, y, z as
$\begin{align}
  & x=\dfrac{k}{\cos \theta }, \\
 & y=\dfrac{k}{\cos \left( {{120}^{\circ }}+\theta \right)} \\
 & z=\dfrac{k}{\cos \left( {{240}^{\circ }}+\theta \right)} \\
\end{align}$
Let us consider the term $xy+yz+zx$. From the above relations, we can write that
$\begin{align}
  & \dfrac{k}{\cos \theta }\times \dfrac{k}{\cos \left( {{120}^{\circ }}+\theta \right)}+\dfrac{k}{\cos \left( {{120}^{\circ }}+\theta \right)}\times \dfrac{k}{\cos \left( {{240}^{\circ }}+\theta \right)}+\dfrac{k}{\cos \left( {{240}^{\circ }}+\theta \right)}\times \dfrac{k}{\cos \theta } \\
 & \\
\end{align}$
We can take ${{k}^{2}}$ common and L.C.M of the whole term.
$\begin{align}
  & {{k}^{2}}\left( \dfrac{1}{\cos \theta \cos \left( {{120}^{\circ }}+\theta \right)}+\dfrac{1}{\cos \left( {{120}^{\circ }}+\theta \right)\cos \left( {{240}^{\circ }}+\theta \right)}+\dfrac{1}{\cos \left( {{240}^{\circ }}+\theta \right)\cos \theta } \right) \\
 & ={{k}^{2}}\left( \dfrac{\cos \theta +\cos \left( {{120}^{\circ }}+\theta \right)\cos \left( {{240}^{\circ }}+\theta \right)}{\cos \theta \cos \left( {{120}^{\circ }}+\theta \right)\cos \left( {{240}^{\circ }}+\theta \right)} \right) \\
 & \\
\end{align}$
Let us consider $\cos \theta +\cos \left( {{120}^{\circ }}+\theta \right)\cos \left( {{240}^{\circ }}+\theta \right)$
Applying the cosine formulae, we get
$\cos \theta +\cos {{120}^{\circ }}\cos \theta -\sin {{120}^{\circ }}\sin \theta +\cos {{240}^{\circ }}\cos \theta -\sin {{240}^{\circ }}\sin \theta $
We know the values
$\begin{align}
  & \cos {{120}^{\circ }}=\cos \left( 90+30 \right)=-\sin 30=-\dfrac{1}{2} \\
 & \sin {{120}^{\circ }}=\sin \left( 90+30 \right)=\cos 30=\dfrac{\sqrt{3}}{2} \\
 & \cos {{240}^{\circ }}=\cos \left( 180+60 \right)=-\cos 60=-\dfrac{1}{2} \\
 & \sin {{240}^{\circ }}=\sin \left( 180+60 \right)=-\sin 60=-\dfrac{\sqrt{3}}{2} \\
\end{align}$
Using these values, we get
$\cos \theta -\dfrac{1}{2}\cos \theta -\dfrac{\sqrt{3}}{2}\sin \theta -\dfrac{1}{2}\cos \theta +\dfrac{\sqrt{3}}{2}\sin \theta =\cos \theta -\cos \theta -\dfrac{\sqrt{3}}{2}\sin \theta +\dfrac{\sqrt{3}}{2}\sin \theta =0$
We got the numerator of the fraction as zero.
Hence we can conclude that
${{k}^{2}}\left( \dfrac{\cos \theta +\cos \left( {{120}^{\circ }}+\theta \right)\cos \left( {{240}^{\circ }}+\theta \right)}{\cos \theta \cos \left( {{120}^{\circ }}+\theta \right)\cos \left( {{240}^{\circ }}+\theta \right)} \right)=0$
$xy+yz+zx=0$.
Hence statement-1 is also correct.
$\therefore $ We can conclude that statement 1 and 2 are correct.

So, the correct answer is “Option C”.

Note: We can use a simpler way to solve the problem by writing the term $\cos \left( {{240}^{\circ }}+\theta \right)$ as $\cos \left( {{360}^{\circ }}-\left( {{240}^{\circ }}+\theta \right) \right)=\cos \left( {{120}^{\circ }}-\theta \right)$. Now the required proof for first and the second statements is the same. By checking any one of them, we can get the required answer.