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A=[1t a n x-t a n x1]a n df(x) is define...

`A=[1t a n x-t a n x1]a n df(x)` is defined as `f(x)=d e tdot(A^T A^(-1))` en the value of `(f(f(f(ff(x))))_` is `(ngeq2)` _________.

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The correct Answer is:
1
`A=[(1,tan x),(-tan x,1)]`
Hence, det. `A=sec^(2)x`
`:.` det `A^(T)=sec^(2)x`
Now `f(x)=` det. `(A^(T)A^(-1))`
`=("det. "A^(T)) ("det. "A^(-1))`
`=("det. "A^(T)) ("det. "A)^(-1)`
`=("det."(A^(T)))/("det. (A)")=1`
Hence, `f(x)=1`.
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