If $\frac{x}{\cos \alpha}=\frac{y}{\cos \left(\frac{2 \pi}{3}-\alpha\right)}=\frac{z}{\cos \left(\frac{2…

If $\frac{x}{\cos \alpha}=\frac{y}{\cos \left(\frac{2 \pi}{3}-\alpha\right)}=\frac{z}{\cos \left(\frac{2 \pi}{3}+\alpha\right)}$, then the value of $(x+y+z)$ is equal to
  1. $\frac{1}{2}$
  2. 0
  3. 1
  4. 2

Solution

Since, it is given that $ \begin{aligned} \frac{x}{\cos \alpha} & =\frac{y}{\cos \left(\frac{2 \pi}{3}-\alpha\right)}=\frac{z}{\cos \left(\frac{2 \pi}{3}+\alpha\right)}=k(\text { let }) \\ \therefore \quad x= & k \cos \alpha, y=k \cos \left(\frac{2 \pi}{3}-\alpha\right) \text { and } \\ z & =k \cos \left(\frac{2 \pi}{3}+\alpha\right) \\ \therefore \quad x+y+z & =k \\ & {\left[\cos \alpha+\cos \left(\frac{2 \pi}{3}-\alpha\right)+\cos \left(\frac{2 \pi}{3}+\alpha\right)\right] } \\ & =k\left[\cos \alpha+2 \cos \frac{2 \pi}{3} \cos \alpha\right] \\ & =k\left[\cos \alpha+2\left(-\frac{1}{2}\right) \cos \alpha\right] \\ & =k[\cos \alpha-\cos \alpha]=0 \end{aligned} $

Asked in: AP EAMCET 2020 (22 Sep Shift 1)

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