If `t a nbeta=(ns inalphacosalpha)/(1-ns in^2alpha)` , show that `tan(alpha-beta)=(1-n)t a nalphadot`

Explanation

`tan beta=(nsin alpha cos alpha)/(1-n sin^(2)alpha)=([(nsinalphacos alpha)/(cos^(2)alpha)])/([(1)/(cos^(2)alpha)-(n sin^(2)alpha)/(cos^(2)alpha)])`
[Dividing numerator and enominator by `cos^(2)alpha`]
`=(n tan alpha)/(sec^(2)alpha-n tan^(2)alpha)=(n tan alpha)/(1+tan^(2)alpha-n tan^(2)alpha)`
`=(n tan alpha)/(1+(1-n)tan^(2)alpha)`
Now, `tan(alpha-beta)=(tan alpha-tan beta)/(1+tan alpha tan beta)`
`=[(tan alpha-(n tan alpha)/(1+(1-n)tan^(2)alpha))/(1+tan alpha(n tan alpha)/(1+(1-n)tan^(2)alpha))]`(From Eq. i)
`=(tan alpha+(1-n)tan^(3)alpha-n tan alpha)/(1+(1-n)(tan^(2)alpha+n tan^(2)alpha)`
`=((1-n)tan alpha+(1-n)tan^(3)alpha)/(1+tan^(2)alpha)`
`=(1-n)tan alpha`