If `alpha,beta "and" gamma` are real number without expanding at any stage prove that
`|{:(1,cos(beta-alpha),cos(gamma-alpha)),(cos(alpha-beta),1,cos(gamma-beta)),(cos(alpha-gamma),cos(beta-gamma),1):}|`=0.

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

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

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

Let D be the determinant given by D = |(1,cos (beta - alpha),cos (gamma - alpha)),(cos (alpha -beta),1,cos (gamma - beta)),(cos (alpha - gamma),cos (beta - gamma),1)| where alpha, beta and gamma are real number Statement -1: The value of D is zero Statement 2: The determinant D is expressible as the product of two determinant each equal to zero

cos alphasin(beta-gamma)+cosbetasin(gamma-alpha)+cos gammasin(alpha-beta)=

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

If alpha, beta, gamma are three real numbers and A=[(1,cos (alpha-beta),cos(alpha-gamma)),(cos (beta-alpha),1,cos (beta-gamma)),(cos (gamma-alpha),cos (gamma-beta),1)] then which of following is/are true ?