Question

lf for three numbers A,B,C,$\sum \left(\frac{A}{B}\right)=1,$ then value of $\cos (\alpha -\beta )+\cos (\beta -\gamma )+\cos (\gamma -\alpha )$ &$\sin (\alpha -\beta )+\sin (\beta -\gamma )+\sin (\gamma -\alpha )$ are respectively given by the ordered pair

• A. $(0,1)$
• B. $(1,0)$
• C. $(\frac{3}{2},\frac{-3}{2})$
• D. $(1,-1)$

Answer 1

Answer: (B)

$(1,0)$

It is given that

$A=e^{i\alpha },B=e^{i\beta },C=e^{i\gamma }$

Where

$i=\sqrt{-1}$

Then,

$\sum \frac{A}{B}=1$

Means,

$\frac{e^{i\alpha }}{e^{i\beta }}+\frac{e^{i\beta }}{e^{i\gamma }}+\frac{e^{i\gamma }}{e^{i\alpha }}=1$

$\Rightarrow e^{i(\alpha -\beta )}+e^{i(\beta -\gamma )}+e^{\gamma -\alpha )}=1$

Breaking,

$e^{i(\alpha -\beta )}=\cos (\alpha -\beta )+i\sin (\alpha -\beta )$

$\Rightarrow \cos (\alpha -\beta )+i\sin (\alpha -\beta )+\cos (\beta -\gamma )+i\sin (\beta -\gamma )+\cos (\gamma -\alpha )+i\sin (\gamma -\alpha )=1$

$\Rightarrow [\cos (\alpha -\beta )+\cos (\beta -\gamma )+\cos (\gamma -\alpha )]+i[\sin (\alpha -\beta )+\sin (\beta -\gamma )+\sin (\gamma -\alpha )]=1+0i$

Equating the real and imaginary parts

$\cos (\alpha -\beta )+\cos (\beta -\gamma )+\cos (\gamma -\alpha )=1$

$\sin (\alpha -\beta )+\sin (\beta -\gamma )+\sin (\gamma -\alpha )=0$

Hence the ordered pair is $(1,0)$