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. $a=1,b=-1$
• B. $a=se{c}^2\theta ,b=0$
• C. $a=0,b=si{n}^2\theta$
• D. $a=sin2\theta ,b=cos2\theta$

Answer 1

The correct option is B $a=se{c}^2\theta ,b=0$

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

$\left[\begin{array}{cc}1+ta{n}^2\theta & tan\theta -tan\theta \\ tan\theta -tan\theta & ta{n}^2\theta +1\end{array}\right]=\left[\begin{array}{cc}a& -b\\ -b& a\end{array}\right]$

$\left[\begin{array}{cc}se{c}^2\theta & 0\\ 0& se{c}^2\theta \end{array}\right]=\left[\begin{array}{cc}a& -b\\ -b& a\end{array}\right]$

Hence, $a=se{c}^2\theta$ and $b=0$.