Question

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

• A. $a=1$
• B. $a={\sec }^2\theta ,b=0$
• C. $a=0,b={\sin }^2\theta$
• D. $a=\sin 2\theta ,b=\cos 2\theta$

Answer 1

Answer: (B)

$a={\sec }^2\theta ,b=0$

L.H.S. $=\left[\begin{array}{cc}1& -tan\theta \\ tan\theta & 1\end{array}\right]\left[\begin{array}{cc}1& tan\theta \\ -tan\theta & 1\end{array}\right]$

$=\left[\begin{array}{cc}(1\times 1)-\left(\tan \theta \right(-\tan \theta \left)\right)& (1\times tan\theta )-(\tan \theta \times 1)\\ (\tan \theta \times 1)+(1\times (-tan\theta \left)\right)& (\tan \theta \times \tan \theta )+(1\times 1)\end{array}\right]$

$=\left[\begin{array}{cc}1+{\tan }^2\theta & 0\\ 0& 1+{\tan }^2\theta \end{array}\right]=\left[\begin{array}{cc}{\sec }^2\theta & 0\\ 0& {\sec }^2\theta \end{array}\right]$

On comparing with R.H.S, we get

$a={\sec }^2\theta ,b=0$