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`|(1,cos(beta-alpha),cos(gamma-alpha)), (cos(alpha-beta),1,cos(gamma-beta)), (cos(beta-alpha), cos(beta - gamma),1)| = `A. `cos alpha cos beta cos gamma`B. `cos alpha + cos beta + cos gamma`C. 1D. 0 |
Answer» Correct Answer - D 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` |
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