Question

If $\left[\begin{array}{cc}\sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{array}\right]\left[\begin{array}{cc}\sin \theta & \cos \theta \\ -\cos \theta & \sin \theta \end{array}\right]=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right],$ then the value of $a+b+c+d$ is

• A. $0$
• B. $\sin \theta +\cos \theta +1$
• C. $2$
• D. $1$

Answer 1

The correct option is C $2$

Given : $\left[\begin{array}{cc}\sin \theta & -\cos \theta \\ \cos \theta & \sin \theta \end{array}\right]\left[\begin{array}{cc}\sin \theta & \cos \theta \\ -\cos \theta & \sin \theta \end{array}\right]=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}{\sin }^2\theta +{\cos }^2\theta & \sin \theta \cos \theta -\cos \theta \sin \theta \\ \sin \theta \cos \theta -\sin \theta \cos \theta & {\sin }^2\theta +{\cos }^2\theta \end{array}\right]=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

$\Rightarrow \left[\begin{array}{cc}1& 0\\ 0& 1\end{array}\right]=\left[\begin{array}{cc}a& b\\ c& d\end{array}\right]$

On comparing the corresponding entries, we have
$a=1$, $b=0$, $c=0$ and $d=1$
$\therefore a+b+c+d=2$