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यदि `tanbeta= ( n sinalpha cos alpha)/(1-n sin^(2)alpha) ` साबित करें कि
`tan ( alpha-beta) = ( 1-n) tanalpha`

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`tan beta = ( n sin alpha cos alpha) /(1-n sin^(2)alpha) = ((nsinalphacosalpha)/(cos^(2)alpha))/((1)/(cos^(2)alpha)-(nsin^(2)alpha)/(cos^(2)alpha))=(ntanalpha)/(sec^(2)alpha-ntan^(2)alpha)`
`= ( n tanalpha)/(1+tan^(2) alpha-ntan^(2) alpha) = ( ntanalpha)/(1+(1-n)tan^(2)alpha)`
अब L.H.S. `=tan(alpha-beta) = ( tanalpha-tanbeta)/(1+tanalphatanbeta)`
`= (tanalpha - ( n tanalpha)/(1+(1-n)tan^(2)alpha))/(1+tanalpha(n tanalpha)/(1+(1-n)tan^(2)alpha))` [(i) से ]
`=(tanalpha+(1-n)tan^(3)alpha-ntanalpha)/(1+(1-n)tan^(2)alpha+ntan^(2)alpha)`
`=((1-n)tanalpha+(1-n)tan^(3)alpha)/(1+tan^(2)alpha)`
`= ((1-n)tan alpha(1+tan^(2)alpha))/(1+tan^(2)alpha)=(1-n)tanalpha`

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