सिद्ध करें की `sin alpha*sin( beta-gamma)+sin beta* sin(gamma-alpha)+sin gamma*sin(alpha-beta)=0`

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

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

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

lf, cos (x + alpha) cos (x + beta), cos (x + gamma) sin (x + alpha), sin (x + beta), sin (x + gamma) sin (beta + gamma), sin ( gamma-alpha), sin (alpha-beta) then (Given alpha! = beta! = gamma), sin (alpha-beta)

Let alpha,beta and gamma are the angles of a right angle triangle,then the value of sin alpha*sin beta*sin(alpha-beta)+sin beta*sin gamma*sin(beta-gamma)+sin gamma*sin alpha*sin(gamma-beta)+sin(beta-gamma)*sin(gamma-gamma)*sin(gamma-

if 0ltalpha,beta,gamma lt(pi/2) prove that sin alpha+sin beta+sin gamma gtsin(alpha+beta+gamma)