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