If `cos^4 theta sec^2 alpha,1/2` and `sin^4 theta cosec^2 alpha` are in `A.P.`, then `cos^8 theta sec^6 alpha ,1/2` and `sin^8 theta cosec^6 alpha` are

Options

A. A.P.
B. G.P.
C. H.P.
D. none of these

Answer

A

Explanation

It is given that
`cos^(4)thetasec^(2)alpha,1/2and sin^(4)thetacosec^(2)alpha"are in"A.P.`
`therefore1=cos^(4)thetasec^(2)alpha+sin^(4)thetacosec^(2)alpha`
`impliescos^(4)thetasin^(2)alpha+sin^(4)thetacos^(2)alpha=sin^(2)alphacos^(2)alpha`
`implies(-1sin^(2)theta)cos^(2)thetasin^(2)alpha+sin^(4)theta(1-sin^(2)theta)=sin^(2)alpha(1-sin^(2)theta)`
`impliescos^(2)thetasin^(2)alpha+sin^(4)theta-sin^(2)thetasin^(2)alpha(cos^(2)theta+sin^(2)theta)=sin^(2)alpha-sin^(4)alpha`
`impliessin^(4)theta+sin^(4)alpha-2sin^(2)thetasin^(2)alpha=2`
`implies(sin^(2)theta-sin^(2)alpha)^(2)=0`
`impliessin^(2)theta=sin^(2)alphaand cos^(2)theta=cos^(2)alpha`
`thereforecos^(8)thetasec^(6)alpha+sin^(8)thetacosec^(6)alpha=cos^(2)theta+sin^(2)theta=1`
`impliescos^(8)thetasec^(6)alpha+sin^(8)thetacosec^(6)alpha "are in"A.P.`