" 0."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 )=

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

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 :

If cos alpha+cos beta+cos gamma=0=sin alpha+sin beta+sin gamma , then (sin3alpha+sin3beta+sin3gamma)/(sin(alpha+beta+gamma)) is equal to

If cos alpha+cos beta+cos gamma=0=sin alpha+sin beta+sin gamma , then (sin3alpha+sin3beta+sin3gamma)/(sin(alpha+beta+gamma)) is equal to

If cos alpha+cos beta+cos gamma=0=sin alpha+sin beta+sin gamma , then (sin3alpha+sin3beta+sin3gamma)/(sin(alpha+beta+gamma)) is equal to

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

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=sin alpha+sin beta+sin gamma then x^(sin(2 alpha-beta-gamma)*x^(sin(2 beta-gamma-alpha)))*x ^(sin(2 gamma-alpha-beta)) is