If 1 - tan θ tan θ 1 1 tan θ - tan θ 1 - 1 = a - b b a , then

Assume,

$A=\left[\begin{array}{cc}1& \tan \theta \\ -\tan \theta & 1\end{array}\right]$

$\left|A\right|=1+{\tan }^2\theta$

and $\mathrm{Adj}\left(A\right)=\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]$

We know, $A^{-1}=\frac{\mathrm{Adj}\left(A\right)}{\left|A\right|}$

Given,

$\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]{\left[\begin{array}{cc}1& \tan \theta \\ -\tan \theta & 1\end{array}\right]}^{-1}=\left[\begin{array}{ll}a& -b\\ b& a\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]\frac{1}{1+{\tan }^2\theta }$ $\left[\begin{array}{cc}1& -\tan \theta \\ \tan \theta & 1\end{array}\right]=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$

$\Rightarrow \frac{1}{1+{\tan }^2\theta }\left[\begin{array}{cc}1-{\tan }^2\theta & -2\tan \theta \\ 2\tan \theta & 1-{\tan }^2\theta \end{array}\right]=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}\frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }& \frac{-2\tan \theta }{1+{\tan }^2\theta }\\ \frac{2\tan \theta }{1+{\tan }^2\theta }& \frac{1-{\tan }^2\theta }{1+{\tan }^2\theta }\end{array}\right]=\left[\begin{array}{cc}a& -b\\ b& a\end{array}\right]$

$\Rightarrow \left[\begin{array}{rr}\cos 2\theta & -\sin 2\theta \\ \sin 2\theta & \cos 2\theta \end{array}\right]=\left[\begin{array}{rr}a& -b\\ b& a\end{array}\right]$

$\Rightarrow a=\cos 2\theta , b=\sin 2\theta$