Question

If $\left[\begin{array}{cc}1& −\tan θ\\ \tan θ& 1\end{array}\right]{\left[\begin{array}{cc}1& \tan θ\\ −\tan θ& 1\end{array}\right]}^{−1}=\left[\begin{array}{cc}a& −b\\ b& a\end{array}\right]$ then

• A. $a=1=b$
• B. $a=\cos 2θ,b=\sin 2θ$
• C. $a=\sin 2θ,b=\cos 2θ$
• D. $a=\cos θ,b=\sin θ$

Answer 1

Answer: (B)

$\left[\begin{array}{cc}1& −\tan θ\\ \tan θ& 1\end{array}\right]$
${\left[\begin{array}{cc}1& \tan θ\\ −\tan θ& 1\end{array}\right]}^{−1}=\left[\begin{array}{cc}a& −b\\ b& a\end{array}\right]$
Since $A^{−1}=\frac{adjA}{âˆ\pounds Aâˆ\pounds }$
$∴\left[\begin{array}{cc}1& −\tan θ\\ \tan θ& 1\end{array}\right]$
$\frac{\left[\begin{array}{cc}1& −\tan θ\\ \tan θ& 1\end{array}\right]}{1+{\tan }^2θ}=\left[\begin{array}{cc}a& −b\\ b& a\end{array}\right]$
$\frac{1}{1+{\tan }^2θ}\left[\begin{array}{cc}1−{\tan }^2θ& −\tan θ−\tan θ\\ \tan θ+\tan θ& −{\tan }^2θ+1\end{array}\right]$
$=\left[\begin{array}{cc}a& −b\\ b& a\end{array}\right]$
$\left[\begin{array}{cc}\frac{1−{\tan }^2θ}{1+{\tan }^2θ}& \frac{−2\tan θ}{1+{\tan }^2θ}\\ \frac{2\tan θ}{1+{\tan }^2θ}& \frac{1−{\tan }^2θ}{1+{\tan }^2θ}\end{array}\right]=\left[\begin{array}{cc}a& −b\\ b& a\end{array}\right]$
$⇒\left[\begin{array}{cc}\cos 2θ& −\sin 2θ\\ \sin 2θ& \cos 2θ\end{array}\right]=\left[\begin{array}{cc}a& −b\\ b& a\end{array}\right]$
So, $a=\cos 2θ,b=\sin 2θ$