Prove that : `cos^2 (beta-gamma) + cos^2 (gamma-alpha) + cos^2 (alpha-beta) =1+2cos (beta-gamma) cos (gamma-alpha) cos (alpha-beta)`.

If cos (beta-gamma)+cos(gamma-alpha)+cos(alpha-beta)=-(3)/(2) then

det [[1, cos (beta-alpha), cos (gamma-alpha) cos (alpha-beta), 1, cos (gamma-beta) cos (beta-alpha), cos (beta-gamma), 1]] =

prove that | (sin alpha, sin beta, sin gamma), (cos alpha, cos beta, cos gamma), (sin 2alpha, sin 2beta, sin 2gamma) | = 4sin (1/2) (beta-gamma) * sin (1/2) (gamma-alpha) * sin (1/2) (alpha-beta) xx {sin (beta + gamma) + sin (gamma + alpha) + sin (alpha + beta)}.

cos alpha sin (beta-gamma) + cos beta sin (gamma-alpha) + cos gamma sin (alpha-beta) =

Let A and B denote the statements A : cos alpha + cos beta + cos gamma =0 B : sin alpha + siin beta + sin gamma = 0 If cos(beta - gamma) + cos (gamma -alpha) + cos (alpha -beta) = - (3)/(2) , then

If alpha,beta,gamma are the angles of a triangle and system of equations cos(alpha-beta)x+cos(beta-gamma)y+cos(gamma-alpha)z=0cos(alpha+beta)x+cos(beta+gamma)y+cos(gamma+alpha)z=0sin(alpha+beta)x+sin(beta+gamma)y+sin(gamma+alpha)z=0 has non-trivial solutions,then triangle is necessarily a.equilateral b.isosceles c.right angled