Question

If $I=\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{- 1}=\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$ , then $\left(a^{2} + b^{2}\right)$ is equal to $\ldots \ldots $

Answer 1

Here, it is given that
$I=\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\begin{bmatrix} 1 & \tan\theta \\ -\tan\theta & 1 \end{bmatrix}^{- 1}$
$=\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}\times \frac{1}{1 + \tan^{2} \theta }\begin{bmatrix} 1 & -\tan\theta \\ \tan\theta & 1 \end{bmatrix}$
$=\frac{1}{1 + \tan^{2} \theta }\begin{bmatrix} 1-\tan^{2}\theta & -2\tan\theta \\ 2\tan\theta & 1-\tan^{2}\theta \end{bmatrix}$
$=\begin{bmatrix} \frac{1 - \tan^{2} \theta }{1 + \tan^{2} \theta } & \frac{- 2 \tan \theta }{1 + \tan^{2} \theta } \\ \frac{2 \tan \theta }{1 + \tan^{2} \theta } & \frac{1 - \tan^{2} \theta }{1 + \tan^{2} \theta } \end{bmatrix}$
$=\begin{bmatrix} \cos 2\theta & -\sin 2\theta \\ \sin2\theta & \cos2\theta \end{bmatrix}$
since, $I=\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$
$\Rightarrow \begin{bmatrix}\cos2\theta & -\sin2\theta \\ \sin2\theta & \cos2\theta \end{bmatrix}=\begin{bmatrix} a & -b \\ b & a \end{bmatrix}$
$\therefore \cos^{2}2\theta +\sin^{2}2\theta =1$