Question

If $a=\cos \alpha +i\sin \alpha$, $b=\cos \beta +i\sin \beta$, $c=\cos \gamma +i\sin \gamma$, 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. $0$
• B. $1$
• C. $-1$
• D. None of these

Answer 1

The correct option is B

$1$

Explanation for the correct option:

Step 1. Find the value of $\cos (\beta -\gamma )+\cos (\gamma -\alpha )+\cos (\alpha -\beta )$:

Given, $a=\cos \alpha +i\sin \alpha$,

$b=\cos \beta +i\sin \beta$,

and $c=\cos \gamma +i\sin \gamma$

Now, $\frac{b}{c}=\frac{\cos \beta +i\sin \beta }{\cos \gamma +i\sin \gamma }\times \frac{\cos \gamma -i\sin \gamma }{\cos \gamma -i\sin \gamma }$

$=\cos \beta \cos \gamma +i\sin \beta \sin \gamma$

$\Rightarrow$ $\frac{b}{c}=\cos (\beta -\gamma )+i\sin (\beta -\gamma )$ …..(1)

Similarly,

$\frac{c}{a}=\cos (\gamma -\alpha )+i\sin (\gamma -\alpha )$ …..(2)

and $\frac{a}{b}=\cos (\alpha -\beta )+i\sin (\alpha -\beta )$ …..(3)

Step 2. Add equation (1), (2) an (3), we get

$\cos (\beta -\alpha )+\cos (\gamma -\alpha )+\cos (\alpha +\beta )+i\left[\sin \right(\beta -\gamma )+\sin (\gamma -\alpha )+\sin (\alpha -\beta \left)\right]=1$ $\left[\mathbf{\because }\frac{\mathbf{b}}{\mathbf{c}}\mathbf{+}\frac{\mathbf{c}}{\mathbf{a}}\mathbf{+}\frac{\mathbf{a}}{\mathbf{b}}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}\right]$

Equate real parts, we get

$\mathbf{\therefore }\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathit{\beta }\mathbf{-}\mathit{\gamma }\mathbf{)}\mathbf{+}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathit{\gamma }\mathbf{-}\mathit{\alpha }\mathbf{)}\mathbf{+}\mathit{c}\mathit{o}\mathit{s}\mathbf{(}\mathit{\alpha }\mathbf{-}\mathit{\beta }\mathbf{)}\mathbf{}\mathbf{=}\mathbf{}\mathbf{1}$

Hence, Option ‘B’ is Correct.