(sin alpha+sin beta+sin gamma)/(cos alpha+cos beta+cos gamma)

If sin alpha+sin beta+sin gamma=0=cos alpha+cos beta+cos gamma, then the value of sin^(2)alpha+sin^(2)beta+sin^(2)gamma, is

If sin alpha + sin beta +sin gamma=0 =cos alpha+cos beta+cos gamma=0 "Show that" sin^2 alpha+sin^2beta+sin^2gamma=cos^2 alpha+ cos^2beta+cos^2gamma=3/2

If cos alpha+ cos beta+ cos gamma + cos alpha cos beta cos gamma=0 , show that sin alpha (1+ cos beta cos gamma)=+- sin beta sin gamma .

lfsin alpha + sin beta + sin gamma = 0 = cos alpha + cos beta + cos gamma then sum cos (alpha-beta) =

If alpha, beta , gamma are angles of a triangle then the value of (sin^(2) alpha + sin ^(2) beta+sin ^(2) gamma-2 cos alpha cos beta cos gamma) is-

If sinalpha+sinbeta+singamma=cosalpha+cosbeta+cosgamma=0, then (A) cos3alpha+cos3beta+cos3gamma=3cos(alpha+beta+gamma) (B) cos 3alpha+cos3beta+cos3gamma=0 (C) sin3alpha+sin3beta+sin3gamma=0 (D) sin3alpha+sin3beta+sin3gamma= 3sin(alpha+beta+gamma)

If cos alpha + cos beta + cos gamma = sin alpha + sin beta + sin gamma = 0 , show that sin 3alpha + sin 3 beta + sin 3gamma = 3 sin (alpha + beta + gamma)

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