Question

If $\frac{x}{\cos \theta }=\frac{y}{\cos \left(\theta -{\displaystyle \frac{2\mathrm{\pi }}{3}}\right)}=\frac{z}{\cos \left(\theta +{\displaystyle \frac{2\mathrm{\pi }}{3}}\right)},$ then x + y + z = ____________.

Answer 1

$$\begin{gathered} \frac{x}{\cos \theta }=\frac{y}{\cos \left(\theta -{\displaystyle \frac{2\mathrm{\pi }}{3}}\right)}=\frac{z}{\cos \left(\theta +{\displaystyle \frac{2\mathrm{\pi }}{3}}\right)}\mathrm{say}=K \\ \mathrm{i}.\mathrm{e}.x=K\cos \theta ,y=K\cos \left(\theta -\frac{2\mathrm{\pi }}{3}\right),z=K\cos \left(\theta +\frac{2\mathrm{\pi }}{3}\right) \\ \therefore x+y+z=K\left[\cos \theta +\cos \left(\theta -\frac{2\mathrm{\pi }}{3}\right)+\cos \left(\theta +\frac{2\mathrm{\pi }}{3}\right)\right] \\ \left[\mathrm{using}\mathrm{identity}:\cos a+\cos b=2\cos \left(\frac{a+b}{2}\right)\cos \left(\frac{a-b}{2}\right)\right] \\ =K\left[\cos \theta +2\cos \theta \cos \left(2\mathrm{\pi }/3\right)\right]\left(\because \frac{\theta -{\displaystyle \frac{2\mathrm{\pi }}{3}}+\theta +{\displaystyle \frac{2\mathrm{\pi }}{3}}}{2}=0\mathrm{and}\cos (-x)=\cos x\right) \\ =K\left[\cos \theta +2\cos \theta \left(-\frac{1}{2}\right)\right]\left[\because \cos \frac{2\mathrm{\pi }}{3}=-\frac{1}{2}\right] \\ =k\left[\cos \theta -\cos \theta \right] \\ =0 \\ \therefore x+y+z=0 \end{gathered}$$