If $A =\begin{bmatrix}1&tanx\\-tanx&1\end{bmatrix}$, then the value of $|A^TA^{-1}|$ is

1

$A^T =\begin{bmatrix}1&-tanx\\tanx&1\end{bmatrix}$

and $A^{-1} =\frac{1}{1+tan^2x}\begin{bmatrix}1&-tanx\\tanx&1\end{bmatrix}$

$⇒A^TA^{-1}\begin{bmatrix}cos2x&-sin2x\\sin2x&cos2x\end{bmatrix}$

Hence (D) is the correct answer.

$⇒|A^TA^{-1}|=1$

Alternatively,

$|A^TA^{-1}| = |A^T| |A^{-1}| = |A| |A^{-1}| = |AA^{-1}| = | I | = 1$