If `A=[[1,tanx],[-tanx,1]],` show that `A^T A^(-1)=[[cos2x,-sin2x],[sin2x,cos2x]]`

Explanation

Here `A = [[1,tanx],[-tanx,1]]`

`:.|A| = [1-tanx(-tanx)] = 1+tan^2x`

`adj. A = [[1,-tanx],[tanx,1]]`

`:.A^-1 = 1/|A| adj.A`

`=>A^-1 = 1/(1+tan^2x) [[1,-tanx],[tanx,1]]`

Now, `A^T = [[1,-tanx],[tanx,1]]`

`:. A^TA^-1 = [[1,-tanx],[tanx,1]]*1/(1+tan^2x) [[1,-tanx],[tanx,1]]`

`= 1/(1+tan^2x) [[1,-tanx],[tanx,1]] [[1,-tanx],[tanx,1]]`

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