[sin(beta+gamma-alpha)+sin(gamma+alpha-beta)],[+sin(alpha+beta-gamma)-sin(alpha+beta+gamma)=]

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)

sin 2 alpha + sin 2 beta + sin 2 gamma - sin2(alpha + beta + gamma )=

"Prove that : " sin alpha + sin beta + sin gamma - sin ( alpha + beta + gamma) =4 "sin" (alpha+beta)/2 "sin"(beta + gamma)/2 "sin"(gamma+alpha)/2

If x=sin(alpha-beta)*sin(gamma-delta), y=sin(beta-gamma)*sin(alpha-delta), z=sin(gamma-alpha) *sin(beta-delta) , then :

If x=sin(alpha-beta)*sin(gamma-delta), y=sin(beta-gamma)*sin(alpha-delta), z=sin(gamma-alpha)*sin(beta-delta) , then :

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 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

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