Question

If $\cos a+\cos b+\cos c=\sin a+\sin b+\sin c=0$ then $\cos \left(a-b\right)+\cos \left(b-c\right)+\cos \left(c-a\right)=-\frac{3}{2}$.

• A. True
• B. False

Answer 1

The correct option is A True

Given, $\cos a+\cos b+\cos c=0$.....(1) and $\sin a+\sin b+\sin c=0$.....(2).

Now squaring (1) and (2) and then adding we get,

${\cos }^{2}a+{\cos }^{2}b+{\cos }^{2}c+2(\cos a.\cos b+\cos b.\cos c+\cos c.\cos a)$ $+{\sin }^{2}a+{\sin }^{2}b+{\sin }^{2}c+2(\sin a.\sin b+\sin b.\sin c+\sin c.\sin a)=0$

or, $2\left(\cos \right(a-b)+\cos (b-c)+\cos (c-a)+3=0$

or, $\cos (a-b)+\cos (b-c)+\cos (c-a)=-\frac{3}{2}$.