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Evaluate `|(cosalphacosbeta,cosalphas inbeta,-sinalpha),(-sinbeta,cosbeta,0),(sinalphacosbeta,sinalphasinbeta,cosalpha)|`

Answer» Let the given determinant be `Delta`. Then,
`Delta = |["cos"alpha"cos"beta, "cos" alpha "sin"beta, -"sin"beta],[-"sin" beta, "cos" beta, 0],["sin"alpha"cos" beta, "sin"alpha "sin"beta, "cos" alpha]|`
`= (1)/("sin"alpha"cos"alpha)*|["sin"alpha"cos"alpha"cos"beta, "sin" alpha "cos"alpha"sin"beta, -"sin"^(2)beta],[-"sin" beta, "cos" beta, 0],["sin"alpha"cos"alpha"cos" beta, "sin"alpha"cos"alpha "sin"beta, "cos"^(2) alpha]|`
`[R_(1) to ("sin" alpha)R_(1), ("cos"alpha)R_(3) " and dividing "Delta "by "("sin" alpha "cos" alpha)]`
`= (1)/("sin"alpha"cos"alpha)*|[0, 0, -1],[-"sin" beta, "cos" beta, 0],["sin"alpha"cos"alpha"cos" beta, "sin"alpha"cos"alpha "sin"beta, "cos"^(2) alpha]|[R_(1) to (R_(1) -R_(3))]`
`= (1)/("sin"alpha"cos"alpha)*(-1)*|[-"sin" beta, "cos" beta],["sin"alpha"cos"alpha"cos" beta, "sin"alpha"cos"alpha "sin"beta]|`
`= (("sin"alpha"cos"alpha))/(("sin" alpha "cos"alpha))*(-1)*|[-"sin" beta, "cos" beta],["cos"beta, "sin"beta]|["taking sin"alpha "cos"alpha "common from"R_(2)]`
`= (-1) *[-"sin"^(2)beta-"cos"^(2) beta]=("sin"^(2) beta + "cos"^(2) beta) =1.`
Hence, `Delta = 1.`


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