Question

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

• A. $0$
• B. $\frac{\pi }{2}$
• C. $\frac{\pi }{4}$
• D. $\frac{\pi }{6}$

Answer 1

The correct option is A $0$

Given, $\left[\begin{array}{cc}1& -tan\theta \\ tan\theta & 1\end{array}\right]{\left[\begin{array}{cc}1& -tan\theta \\ tan\theta & 1\end{array}\right]}^{-1}={\left[\begin{array}{cc}cosa& -sina\\ sina& cosa\end{array}\right]}^{-1}$
Let $A=\left[\begin{array}{cc}1& -tan\theta \\ tan\theta & 1\end{array}\right],B=\left[\begin{array}{cc}cosa& -sina\\ sina& cosa\end{array}\right]$
So, $A{A}^{-1}=B^{-1}$
$\Rightarrow I=B^{-1}$
Here, $|B|=1$
$adjB=C^T={\left[\begin{array}{cc}cosa& -sina\\ sina& cosa\end{array}\right]}^T$
$\Rightarrow adjB=\left[\begin{array}{cc}cosa& sina\\ -sina& cosa\end{array}\right]$
Hence, $B^{-1}=\left[\begin{array}{cc}cosa& sina\\ -sina& cosa\end{array}\right]$
$\Rightarrow \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}cosa& sina\\ -sina& cosa\end{array}\right]$
$\Rightarrow \cos a=1,\sin a=0$
$\Rightarrow a=0$