If the matrix A= (1 Tanx -Tanx 1) Show that A Transpose x A inverse=(Cos2x -Sin2x Sin2x Cos2x) ​

Dear Student,
$A=\left[\begin{array}{cc}1& \mathrm{tan}x\\ -\mathrm{tan}x& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}{A}^{T}=\left[\begin{array}{cc}1& -\mathrm{tan}x\\ \mathrm{tan}x& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}\left|A\right|=1+{\mathrm{tan}}^{2}x=se{c}^{2}x.\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{A}^{-1}=\frac{1}{\left|A\right|}\left[\begin{array}{cc}1& -\mathrm{tan}x\\ \mathrm{tan}x& 1\end{array}\right]=\frac{1}{se{c}^{2}x}\left[\begin{array}{cc}1& -\mathrm{tan}x\\ \mathrm{tan}x& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{A}^{T}×{A}^{-1}=\frac{1}{se{c}^{2}x}\left[\begin{array}{cc}1& -\mathrm{tan}x\\ \mathrm{tan}x& 1\end{array}\right]×\left[\begin{array}{cc}1& -\mathrm{tan}x\\ \mathrm{tan}x& 1\end{array}\right]\phantom{\rule{0ex}{0ex}}=\frac{1}{se{c}^{2}x}\left[\begin{array}{cc}1-{\mathrm{tan}}^{2}x& -2\mathrm{tan}x\\ 2\mathrm{tan}x& 1-{\mathrm{tan}}^{2}x\end{array}\right]=\left[\begin{array}{cc}\frac{1-{\mathrm{tan}}^{2}x}{se{c}^{2}x}& \frac{-2\mathrm{tan}x}{se{c}^{2}x}\\ \frac{2\mathrm{tan}x}{se{c}^{2}x}& \frac{1-{\mathrm{tan}}^{2}x}{se{c}^{2}x}\end{array}\right]=\left[\begin{array}{cc}\frac{{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}xse{c}^{2}x}& \frac{-2\mathrm{tan}x}{se{c}^{2}x}\\ \frac{2\mathrm{tan}x}{se{c}^{2}x}& \frac{{\mathrm{cos}}^{2}x-{\mathrm{sin}}^{2}x}{{\mathrm{cos}}^{2}xse{c}^{2}x}\end{array}\right]=\left[\begin{array}{cc}\mathrm{cos}2x& -2\mathrm{sin}x\mathrm{cos}x\\ 2\mathrm{sin}x\mathrm{cos}x& \mathrm{cos}2x\end{array}\right]=\left[\begin{array}{cc}\mathrm{cos}2x& -\mathrm{sin}2x\\ \mathrm{sin}2x& \mathrm{cos}2x\end{array}\right]=RHS$

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