Q69 / 120 · VITEEE 2014 · Mathematics
If $\cos \alpha+i \sin \alpha, b=\cos \beta+i \sin \beta$, $c=\cos \gamma+i \sin \gamma$ and $\frac{b}{c}+\frac{c}{a}+\frac{a}{b}=1$, then $\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)$ is equal to
- A$\frac{3}{2}$
- B$-\frac{3}{2}$
- C0
- Correct answerD1
Solution
Given: $a=\cos \alpha+i \sin \alpha$ $b=\cos \beta+i \sin \beta$
and $\mathrm{c}=\cos \gamma+\mathrm{i} \sin \gamma$
Now, $\frac{\mathrm{b}}{\mathrm{c}}=\frac{\cos \beta+\mathrm{i} \sin \beta}{\cos \gamma+\mathrm{i} \sin \gamma} \times \frac{\cos \gamma-\mathrm{i} \sin \gamma}{\cos \gamma-\mathrm{i} \sin \gamma}$
$\begin{aligned} \cos \beta \cdot \cos \gamma+\sin \beta \cdot \sin \gamma+\mathrm{i} & \\ &[\sin \beta \cdot \cos \gamma-\sin \gamma \cdot \cos \beta] \end{aligned}$
$\Rightarrow \frac{\mathrm{b}}{\mathrm{c}}=\cos (\beta-\gamma)+\mathrm{i} \sin (\beta-\gamma)$
Similarly, $\frac{\mathrm{c}}{\mathrm{a}}=\cos (\gamma-\alpha)+\mathrm{i} \sin (\gamma-\alpha)$ and $\frac{\mathrm{a}}{\mathrm{b}}=\cos (\alpha-\beta)+\mathrm{i} \sin (\alpha-\beta)$
On adding Eqs. (i), (ii) and (iii), we get $\cos (\beta-\alpha)+\cos (\gamma-\alpha)+\cos (\alpha+\beta)+i$
$\begin{array}{r}
{[\sin (\beta-\gamma)+\sin (\gamma-\alpha)+\sin (\alpha-\beta)]=1} \\
{\left[\because \frac{\mathrm{b}}{\mathrm{c}}+\frac{\mathrm{c}}{\mathrm{a}}+\frac{\mathrm{a}}{\mathrm{b}}=1\right]}
\end{array}$
On equating real parts, we get $\cos (\beta-\gamma)+\cos (\gamma-\alpha)+\cos (\alpha-\beta)=1$
Hence, the correct answer is option: 'D'.