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`

Options

A. `cos alpha cos beta cos gamma`
B. `cos alpha + cos beta + cos gamma`
C. 1
D. 0

Answer

D

Explanation

We have,
`= Delta= |(1,cos (beta- alpha),cos (gamma - alpha)),(cos (alpha - beta),1,cos (gamma - beta)),(cos (alpha - gamma),cos (beta - gamma),1)|`
`= |(cos^(2) alpha + sin^(2) alpha,cos beta cos alpha + sin beta alpha,cos alpha cos gamma + sin alpha sin gamma),(cos alpha cos beta + sin alpah sin beta,"cos"^(2)beta + sin^(2) beta,cos beta cos gamma + sin beta sin gamma),(cos alpha cos gamma + sin alpha sin gamma,cos beta cos gamma + sin beta sin gamma,cos^(2) beta + sin^(2) beta)|`
`= |(cos alpha,sin alpha,0),(cos beta,sin beta,0),(cos gamma,sin gamma,0)|.|(cos alpha,sin alpha,0),(cos beta,sin beta,0),(cos gamma,sin gamma,0)| = 0`