Question

If $a=cos\alpha +isin\alpha ,b=cos\beta +isin\beta ,c=cos\gamma +isin\gamma$ 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. $-\frac{3}{2}$
• C. $0$
• D. $1$

Answer 1

Answer: (D)

$1$

Given : $a=cos\alpha +isin\alpha$
$b=cos\beta +isin\beta$
and $c=cos\gamma +isin\gamma$
Now, $\frac{b}{c}=\frac{cos\beta +isin\beta }{cos\gamma +isin\gamma }\times \frac{cos\gamma -isin\gamma }{cos\gamma -isin\gamma }$
$=cos\beta cos\gamma +sin\beta sin\gamma +i$$[sin\beta cos\gamma -sin\gamma cos\beta ]$
$⟹\frac{b}{c}=cos(\beta -\gamma )+isin(\beta -\gamma )$ ...... (i)
Similarly, $\frac{c}{a}=cos(\gamma -\alpha )+isin(\gamma -\alpha )$ ....... (ii)
and $\frac{a}{b}=cos(\alpha -\beta )+isin(\alpha -\beta )$ ....... (iii)
On adding Eqs. (i), (ii) and (iii), we get
$cos(\beta -\alpha )+cos(\gamma -\alpha )+cos(\alpha +\beta )+i$$\left[sin\right(\beta -\gamma )+sin(\gamma -\alpha )+sin(\alpha -\beta \left)\right]=1$ $[\because \frac{b}{c}+\frac{c}{a}+\frac{a}{b}=1]$
On equating real parts, we get
$cos(\beta -\gamma )+cos(\gamma -\alpha )+cos(\alpha -\beta )=1$

Hence, option D is correct.