1.

If `cos(theta - alpha) , costheta , cos(theta + alpha)` are in H.P. then `costheta.sec(alpha)/2 = `

Answer» If three numbers `a,b and c` are in H.P., then,
`b = (2ac)/(a+c)`
As, `cos(theta-alpha),costheta,cos(theta+alpha)` are in H.P.,
`:. costheta = (2cos(theta-alpha)cos(theta+alpha))/(cos(theta-alpha)+cos(theta+alpha))`
`=>costheta = (2(cos^2thetacos^2alpha - sin^2thetasin^2alpha))/(2costhetacosalpha)`
`=>costheta = (cos^2thetacos^2alpha - sin^2thetasin^2alpha)/(costhetacosalpha)`
`=>costheta = (cos^2thetacos^2alpha - (1-cos^2theta)sin^2alpha)/(costhetacosalpha)`
`=>costheta = (cos^2thetacos^2alpha - sin^2alpha+sin^2alphacos^2theta)/(costhetacosalpha)`
`=>costheta = (cos^2theta(cos^2alpha + sin^2alpha)-sin^2alpha)/(costhetacosalpha)`
`=>cos^2thetacosalpha = cos^2theta-sin^2alpha`
`=>sin^2alpha = cos^2theta(1-cosalpha)`
`=>(2sin(alpha/2)cos(alpha/2))^2 = cos^2theta(2sin^2(alpha/2))`
`=>4sin^2(alpha/2)cos^2(alpha/2) = 2cos^2thetasin^2(alpha/2)`
`=>2cos^2(alpha/2) = cos^2theta`
`=>2 = cos^2thetasec^2(alpha/2)`
`=>costhetasec(alpha/2) = sqrt2`


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