Question

If $x\cos \theta =y\cos (\theta +\frac{\pi }{3})=z\sin (\theta -\frac{\pi }{6})$, where $y\ne 0$, then which of the following is/are correct?

• A. $y=z\forall \theta \in R$
• B. $y+z=0\forall \theta \in R$
• C. If $x=0$, then $\theta$ lies in the first and third quadrant.
• D. There is only one value of $x$ satisfying the equation $x+y+z=0$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $y+z=0\forall \theta \in R$
C If $x=0$, then $\theta$ lies in the first and third quadrant.
D There is only one value of $x$ satisfying the equation $x+y+z=0$

Let $x\cos \theta =y\cos (\theta +\frac{\pi }{3})=z\sin (\theta -\frac{\pi }{6})=k$
$x=\frac{k}{\cos \theta }\cdots \left(1\right)y=\frac{k}{\cos (\theta +\frac{\pi }{3})}\Rightarrow y=\frac{2k}{\cos \theta -\sqrt{3}\sin \theta }\cdots \left(2\right)z=\frac{k}{\sin (\theta -\frac{\pi }{6})}\Rightarrow z=\frac{2k}{\sqrt{3}\sin \theta -\cos \theta }\cdots \left(3\right)$

Now, from equations $\left(2\right)$ and $\left(3\right)$,
$y=-z\Rightarrow y+z=0\cdots \left(4\right)$

When $x=0$,
$x\cos \theta =y\cos (\theta +\frac{\pi }{3})\Rightarrow y\cos (\theta +\frac{\pi }{3})=0\Rightarrow \cos (\theta +\frac{\pi }{3})=0(\because y\ne 0)\Rightarrow \theta +\frac{\pi }{3}=\frac{\pi }{2},\frac{3\pi }{2}\Rightarrow \theta =\frac{\pi }{6},\frac{7\pi }{6}$
So, $x=0$, then $\theta$ lies in first and third quadrant.
If $x+y+z=0$,
$\because y+z=0\Rightarrow x=0$
Therefore there is only one value of $x$ satisfying the equation.