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

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

If cos (alpha + beta) sin (gamma + delta) = cos (alpha - beta) sin (gamma - delta) then:

Prove that: cos alpha+cos beta+cos gamma+cos(alpha+beta+gamma)=4(cos( alpha + beta ))/(2)(cos( beta + gamma ))/(2)(cos( gamma + alpha ))/(2)

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

If cos alpha+cos beta+cos gamma=sin alpha+sin beta+sin gamma=0 , and cos(alpha-beta)+cos(beta-gamma)+cos(gamma-alpha)=6K , K in R , then K is equal to :

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

If origin is the orthocentre of the triangle with vertices A(cos alpha,sin alpha),B(cos beta,sin beta),C(cos gamma,sin gamma) then cos(2 alpha-beta-gamma)+cos(2 beta-gamma-alpha)+cos(2 gamma-alpha-beta)=

If cos alpha+cos beta+cos gamma=0 and also sin alpha+sin beta+sin gamma=0 then prove that

Prove that cos alpha+cos beta+cos gamma+cos (alpha+beta+gamma) =4cos((alpha+beta)/2)cos((beta+gamma)/2)cos((gamma+alpha)/2)