# Solve this

Question:

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

(a) $a=1, b=1$

(b) $a=\cos 2 \theta, b=\sin 2 \theta$

(c) $a=\sin 2 \theta, b=\cos 2 \theta$

(d) none of these

Solution:

(b) $a=\cos 2 \theta, b=\sin 2 \theta$

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

Given :

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

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

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

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

On comparing, we get

$a=\frac{1-\tan ^{2} \theta}{\sec ^{2} \theta}$ and $b=\frac{2 \tan \theta}{\sec ^{2} \theta}$

$\Rightarrow a=\cos ^{2} \theta-\sin ^{2} \theta$ and $b=2 \sin \theta \cos \theta$

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