`xtan(theta - (pi)/6) = ytan(theta + (2pi)/3) ` then `(x +y)/(x-y) ` =

x tan (theta - (pi)/(6)) = y tan (theta + (2pi)/(3)) implies (x + y)/(x -y) =

x tan ( theta - (pi)/(6)) = y tan (theta + ( 2pi)/(3)) implies (x + y)/( x -y) =

If x cos theta = y cos (theta + (2pi)/(3)) = z cos (theta + (4pi)/(3)) , find the value of xy + yz + zx

If x sin theta = y sin ( theta + (2pi)/3) = z sin ( theta+ (4pi)/3), then the value of xy+yz+zx will be-

If xcostheta=ycos(theta+(2pi)/3)=z cos(theta+(4pi)/3) , prove that x y+y z+z x=0.

If xcostheta=ycos(theta+(2pi)/3)=z cos(theta+(4pi)/3) , prove that x y+y z+z x=0.

If quad sin (theta + (7 pi) / (12)) + sin (theta- (pi) / (12)) + sin (theta + (3 pi) / (12)), Y = cos (theta + (7 pi) / (12)) + cos (theta- (pi) / (12)) + cos (theta + (3 pi) / (12)) then prove that (X) / (Y) - (Y) / (X) = 2tan2 theta

If ytan(θ− π/6)=xtan(θ+ 2π/3) , then (y+x)/(y-x) =