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

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

MCQ

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
$a=1=b$
✓
$a=\cos 2 \theta, b=\sin 2 \theta$
C
$a=\sin 2 \theta, b=\cos \theta$
D
$a=\cos \theta, b=\sin \theta$

✓

Answer

Correct option: B.

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

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