If ` A B C` having vertices `A(acostheta_1,asintheta_1),B(acostheta_2asintheta_2),a n dC(acostheta_3,asintheta_3)` is equilateral, then prove that `costheta_1+costheta_2+costheta_3=sintheta_1+sintheta_2+sintheta_3=0.`

Options

A. `costheta_1+costheta_2+costheta_3=0`
B. `sintheta_1+sintheta_2+sin theta_3=0`
C. `tantheta_1+tantheta_2+tantheta_3=0`
D. `cottheta_1+cottheta_2+cottheta_3=0`

Answer

A, B

Explanation

Vertices `(a cos theta_1,a sintheta_1),(acostheta_2,a sintheta_2)`, and origin is the circumcenter (centroid) of circumcircle. Therefore, the coordinates of the centroid are
`((a(costheta_1+costheta_2+costheta_3))/(3),(a(sintheta_1,+sintheta_2+sintheta_3))/(3))`
But as the centroid is the origin (0,0) we have `cos theta_1+costheta_2+costheta_3=0`
and `sin theta_1+sintheta_2+sintheta_3=0`