Without expanding, prove the following
`|(1,1,1),(alpha,beta,gamma),(betagamma,gammaalpha,alphabeta)|=(beta-alpha)(gamma-alpha)(alpha-beta)`

Using properties of determinants, prove that: |[alpha,alpha^2,beta+gamma],[beta,beta^2,gamma+alpha],[gamma,gamma^2,alpha+beta]| = (beta-gamma)(gamma-alpha)(alpha-beta)(alpha+beta+gamma)

Without expanding, show that the following determinants vanish: {:|(1,cos(beta-alpha),cos(gamma-alpha)),(cos(alpha-beta),1,cos(gamma-beta)),(cos(alpha-gamma)),cos(beta-gamma),1|

Prove that |[[alpha,beta,gamma],[alpha^2,beta^2,gamma^2],[alpha^3,beta^3,gamma^3]]|= alpha beta gamma (alpha-beta)(beta-gamma)(gamma-alpha)

If alpha, beta, gamma are the roots of the cubic x^(3)-px^(2)+qx-r=0 Find the equations whose roots are (i) beta gamma +1/(alpha), gamma alpha+1/(beta), alpha beta+1/(gamma) (ii) (beta+gamma-alpha),(gamma+alpha-beta),(alpha+beta-gamma) Also find the valueof (beta+gamma-alpha)(gamma+alpha-beta)(alpha+beta-gamma)

If cosalpha+cosbeta+cosgamma=0=sinalpha+sinbeta+singamma , then which of the following is/are true:- (a) cos(alpha-beta)+cos(beta-gamma)+cos(gamma-delta)=-3/2 (b) cos(alpha-beta)+cos(beta-gamma)+cos(gamma-delta)=-1/2 (c) sumcos2alpha+2cos(alpha+beta)+2cos(beta+gamma)+2cos(gamma+alpha)=0 (d) sumsin2alpha+2sin(alpha+beta)+2sin(beta+gamma)+2sin(gamma+alpha)=0

The value of |{:(betagamma,betagamma'+beta'gamma,beta'gamma'),(gammaalpha,gammaalpha'+gamma'alpha,gamma'alpha'),(alphabeta,alphabeta'+alpha'beta,alpha'beta'):}| is

Prove that |{:(2,,alpha+beta+gamma+delta,,alphabeta+gammadelta),(alpha+beta+gamma+delta,,2(alpha+beta)(gamma+delta),,alphabeta(gamma+delta)+gammadelta(alpha+beta)),(alphabeta+gammadelta,,alphabeta(gamma+delta)+gammadelta(alpha+beta),,2alphabetagammadelta):}|=0

If alpha,beta,gamma are real numbers, then without expanding at any stage, show that |1cos(beta-alpha)"cos"(gamma-alpha)"cos"(alpha-beta)1"cos"(gamma-beta)"cos"(alpha-gamma)"cos"(beta-gamma)1|=0

Prove that cosalpha+cosbeta+cosgamma+cos(alpha+beta+gamma)=4cos(alpha+beta)/(2)cos(beta+gamma)/2cos(gamma+alpha)/2

If alpha+beta=pi/2a n dbeta+gamma=alpha, then tanalpha equals