Let \(\alpha, \beta\) and \(\gamma\) be such that \(0 < \alpha < \beta < \gamma < 2 \pi\). For any \(x \in…
Let \(\alpha, \beta\) and \(\gamma\) be such that \(0 < \alpha < \beta < \gamma < 2 \pi\).
For any \(x \in R\) if \(\cos (x+\alpha)+\cos (x+\beta)+\cos (x+\gamma)=0 \text {, }\)
then \(\tan (\gamma-\alpha)=\)
\(-\sqrt{3}\)
0
1
\(\sqrt{3}\)
Solution
It is given that for any \(x \in R\).
\(\cos (x+\alpha)+\cos (x+\beta)+\cos (x+\gamma)=0\)
Now, put \(x=-\alpha-\beta-\gamma\),
\(0 < \alpha < \beta < \gamma < 2 \pi\), we get
\(\begin{aligned}
& \cos (\beta+\gamma)+\cos (\alpha+\gamma)+\cos (\alpha+\beta)=0 \\
& \Rightarrow 2 \cos \left(\frac{\alpha+2 \beta+\gamma}{2}\right) \cos \left(\frac{\gamma-\alpha}{2}\right) +\cos (\alpha+\gamma)=0 \quad \ldots (i)
\end{aligned}\)
Now, put \(x=\frac{\pi}{2}-\beta\), we get
\(\begin{array}{ll}
& \sin (\beta-\alpha)+\sin (\beta-\gamma)=0 \\
\Rightarrow & \sin (\beta-\alpha)=\sin (\gamma-\beta) \\
\Rightarrow \quad & 2 \beta=\alpha+\gamma \quad \ldots (ii)
\end{array}\)
From Eqs. (i) and (ii), we get
\(\begin{array}{rc}
& \cos (\alpha+\gamma)\left[\frac{1}{2} \cos \left(\frac{\gamma-\alpha}{2}\right)+1\right]=0 \\
\Rightarrow \quad & \quad \cos \left(\frac{\gamma-\alpha}{2}\right)=-\frac{1}{2} \\
\Rightarrow \quad & \frac{\gamma-\alpha}{2}=\frac{2 \pi}{3} \Rightarrow \gamma-\alpha=\frac{\pi}{3}
\end{array}\)
So, \(\tan (\gamma-\alpha)=\sqrt{3}\)