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

Explanation

Given that, `A = [[1,tanx],[-tanx,1]]`

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


`adj(A) = [[1,-tanx],[tanx,1]]`

`therefore A^-1 = 1/|A| adj(A)`

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


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


`therefore 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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