`cos(alpha + beta)cosgamma - cos(beta+gamma)cosalpha = sinbetasin(gamma - alpha)`

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

The expression cos^(2)(alpha + beta + gamma) + cos^(2)(beta + gamma) + cos^(2)alpha - 2cos alpha cos(beta + gamma)cos(alpha + beta + gamma) is (a) independent of alpha (b) independent of beta (c) dependent on gamma only (d) dependent on alpha, beta, gamma

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

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

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

The expression cos^(2)(alpha + beta + gamma) + cos^(2)(beta + gamma) + cos^(2)alpha - 2cos alpha cos(beta + gamma)cos(alpha + beta + gamma) is

If cos (alpha - beta) + cos (beta - gamma) + cos (gamma - alpha) = (-3)/(2) then prove that cos alpha + cos beta+ cos gamma = sin alpha + sin beta + sin gamma = 0 .

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

For all value of alpha,beta,gamma prove that ,cosalpha+cosbeta+cosgamma+cos(alpha+beta+gamma)=4cos(alpha+beta)/2*cos(beta+gamma)/2*cos(gamma+alpha)/2