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Question:
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$\text { If }\left[\begin{array}{cc}
1 & -\tan \theta \\
\tan \theta & 1
\end{array}\right]\left[\begin{array}{cc}
1 & \tan \theta \\
-\tan \theta & 1
\end{array}\right]^{-1}=\left[\begin{array}{cc}
a & -b \\
b & a
\end{array}\right],$
then
Options:
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
Solution:
1746 Upvotes
Verified Answer
The correct answer is:
$a=\cos 2 \theta, b=\sin 2 \theta$
$a=\cos 2 \theta, b=\sin 2 \theta$
Given,
$\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]\left[\begin{array}{cc}1 & \tan \theta \\ -\tan \theta & 1\end{array}\right]^{-1}=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right] \cdot \frac{1}{1+\tan ^2 \theta}\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]$
$=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$
$\Rightarrow \frac{1}{1+\tan ^2 \theta}\left[\begin{array}{cc}1-\tan ^2 \theta & -2 \tan \theta \\ 2 \tan \theta & 1-\tan ^2 \theta\end{array}\right]=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$
$\Rightarrow \quad\left[\begin{array}{cc}\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta} & \frac{2 \tan \theta}{1+\tan ^2 \theta} \\ \frac{2 \tan \theta}{1+\tan ^2 \theta} & \frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\end{array}\right]=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right]=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$
$\Rightarrow \quad a=\cos 2 \theta, b=\sin 2 \theta$
$\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]\left[\begin{array}{cc}1 & \tan \theta \\ -\tan \theta & 1\end{array}\right]^{-1}=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right] \cdot \frac{1}{1+\tan ^2 \theta}\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]$
$=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$
$\Rightarrow \frac{1}{1+\tan ^2 \theta}\left[\begin{array}{cc}1-\tan ^2 \theta & -2 \tan \theta \\ 2 \tan \theta & 1-\tan ^2 \theta\end{array}\right]=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$
$\Rightarrow \quad\left[\begin{array}{cc}\frac{1-\tan ^2 \theta}{1+\tan ^2 \theta} & \frac{2 \tan \theta}{1+\tan ^2 \theta} \\ \frac{2 \tan \theta}{1+\tan ^2 \theta} & \frac{1-\tan ^2 \theta}{1+\tan ^2 \theta}\end{array}\right]=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$
$\Rightarrow\left[\begin{array}{cc}\cos 2 \theta & -\sin 2 \theta \\ \sin 2 \theta & \cos 2 \theta\end{array}\right]=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right]$
$\Rightarrow \quad a=\cos 2 \theta, b=\sin 2 \theta$
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