Explain the equation:- `sin^2 gamma + cos^2 gamma=1`.

The angles alpha , beta , gamma of a triangle satisfy the equations 2 sin alpha + 3 cos beta = 3sqrt(2) and 3sin beta + 2cos alpha = 1 . Then angle gamma equals

If alpha+gamma=2beta then the expression (sin alpha-sin gamma)/(cos gamma-cos alpha) simplifies to:

Prove that : cos^2 (beta-gamma) + cos^2 (gamma-alpha) + cos^2 (alpha-beta) =1+2cos (beta-gamma) cos (gamma-alpha) cos (alpha-beta) .

Suppose alpha ,beta , gamma in R are such that sin alpha, sin beta , sin gamma ne 0 and Delta = |{:(sin^(2) alpha , sin alpha cos alpha , cos^(2) alpha),(sin^(2) beta , sin beta cos beta , cos^(2) beta),(sin^(2) gamma , sin gamma cos gamma , cos^(2) gamma):}| then Delta cannot exceed

Prove that det [[sin alpha, cos alpha, sin (alpha + delta) sin beta, cos beta, sin (beta + delta) sin gamma, cos gamma, sin (gamma + delta)]] = 0

If alpha + beta- gamma= pi , prove that sin^(2)alpha + sin^(2)beta - sin^(2)gamma = 2sin alpha sin beta cos gamma .

prove that | (sin alpha, sin beta, sin gamma), (cos alpha, cos beta, cos gamma), (sin 2alpha, sin 2beta, sin 2gamma) | = 4sin (1/2) (beta-gamma) * sin (1/2) (gamma-alpha) * sin (1/2) (alpha-beta) xx {sin (beta + gamma) + sin (gamma + alpha) + sin (alpha + beta)}.