If [ cc 1 & - & 1 ] [ cc 1 & - & 1 ]^ -1 = [ cc a & -b b & a ], then

We have, [ cc 1 & - & 1 ] [ cc 1 & - & 1 ]^ -1 = [ cc a & -b b & a ] Now, [ cc 1 & - & 1 ]^ -1 =1 ^2 [ cc 1 & - & 1 ] = [ cc ^2 & - ^2 ^2 & ^2 ] [ cc 1 & - & 1 ] [ cc ^2 & - ^2 ^2 & ^2 ]= [ cc a & -b b & a ] [ cc ^2 - ^2 ^2 & -2 ^2 2 ^2 & ^2 - ^2 ^2 ] & = [ cc a & -b b & a ] a= ^2 - ^2 ^2 and b=2 ^2 ^2 & a= ^2 (1- ^2 ^2 ) and b=2 ^2 a= ^2 - ^2 = 2 and b=2 = 2

If [ cc 1 & - & 1 ] [ cc 1 & - & 1 ]^ -1 = [ cc a & -b b & a ], t…

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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

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Answer

We have,
$ \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]$
Now, $\left[\begin{array}{cc}1 & \tan \theta \\ -\tan \theta & 1\end{array}\right]^{-1}=\frac{1}{\sec ^2 \theta}\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]$
$=\left[\begin{array}{cc}\cos ^2 \theta & -\tan \theta \cos ^2 \theta \\ \cos ^2 \theta \tan \theta & \cos ^2 \theta\end{array}\right]$
$ \Rightarrow\left[\begin{array}{cc}1 & -\tan \theta \\ \tan \theta & 1\end{array}\right]\left[\begin{array}{cc}\cos ^2 \theta & -\tan \theta \cos ^2 \theta \\ \cos ^2 \theta \tan \theta & \cos ^2 \theta\end{array}\right]=\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right] $
$ \Rightarrow\left[\begin{array}{cc}\cos ^2 \theta-\cos ^2 \theta \tan ^2 \theta & -2 \tan \theta \cos ^2 \theta \\ 2 \tan \theta \cos ^2 \theta & \cos ^2 \theta-\cos ^2 \theta \tan ^2 \theta\end{array}\right]$
$\begin{aligned} & =\left[\begin{array}{cc}a & -b \\ b & a\end{array}\right] \end{aligned} $
$ \therefore a=\cos ^2 \theta-\cos ^2 \theta \tan ^2 \theta \text { and } b=2 \tan ^2 \cos ^2 \theta$
$\begin{aligned} & \Rightarrow a=\cos ^2 \theta\left(1-\frac{\sin ^2 \theta}{\cos ^2 \theta}\right) \text { and } b=\frac{2 \sin \theta}{\cos \theta} \cdot \cos ^2 \theta \end{aligned} $
$ \Rightarrow a=\cos ^2 \theta-\sin ^2 \theta=\cos 2 \theta \text { and } b=2 \sin \theta \cos \theta=\sin 2 \theta$