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If (cosalpha)/(cosbeta)=m and (cosalpha)...

If `(cosalpha)/(cosbeta)=m` and `(cosalpha)/(sinbeta)=n` show that `(m^2+n^2)cos^2beta=n^2`

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Given
`(cosalpha)/(cosbeta)=m`
R.H.S
`(m^2+n^2)cos^2beta`
=`((cos^2alpha)/(cos^2beta)+(cos^2alpha)/(sin^2beta))cos^2beta`
=`(cos^2alphasin^2beta+cos^2alphacos^2beta)/(sin^2beta)`
=`cos^2alpha(sin^2beta+cos^2beta)/(sin^2beta)`
=`cos^2alpha/sin^2beta`
...
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