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If alpha,beta,gamma are real numbers,...

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`

A

Statement 1 is true, Statement 2 is true, Statement 2 is a correct explanation for Statement 2

B

Statement 1 is true, Statement 2 is true, Statement 2 is not a correct explanation for Statement 2

C

Statement 1 is true, Statement 2 is False

D

Statement 1 is False, Statement 2 is true

Text Solution

Verified by Experts

The correct Answer is:
A
We have,
`D = |(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 sin alpha,cos gamma cos alpha + sin gamma sin alpha),(cos alpha cos beta + sin alpha sin beta,cos^(2) beta + sin^(2) beta,cos gamma cos beta + sin gamma sin beta),(cos alpha cos gamma + sin alpha sin gamma,cos beta cos gamma + sin beta sin gamma,cos^(2) gamma + sin^(2) gamma)|`
`= |(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`
So, both the statement are true and statement 2 is a correct explanation for statement 1
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|[1,cos(alpha-beta), cos alpha] , [cos(alpha-beta),1,cos beta] , [cos alpha, cos beta, 1]|