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.`

Answer

NA

Explanation

Since the circumcenter is at the origin, the orthocenter is
`x_1+x_2+x_3,x_1tantheta_1+x_2tantheta_2+x_3tantheta_3)`
`therefore a=x_1 +x_2+x_3`
and `b=x_1tantheta_1+x_2tantheta_2+x_3tantheta_3`
Also `x_1^(2)+x_1^(2) tantheta_1^(2) =x_2^(2)+x_2^(2)tan theta_2^(2)=x_3^(2)=x_3^(2)tan theta_3^(2)`
or `x_1sec theta_1=x_2sec theta_2=x_3sec theta_3=lambda` (say)
Now, `(a)/(b) =(x_1+x_2+x_3)/(x_1tantheta_1+x_2tan theta_2+x_3tantheta_3)`
`=(lambdacostheta_1+lambdatheta_2+lambdatheta_3)/(lambdacostheta_1thantheta_1+ lambdacostheta_2tantheta_2+lambdacostheta_2tan theta_3)`
`=(costheta_1+costheta_2+costheta_3)/(sintheta_1+sintheta+sintheta_3)` .