Question

The inverse of $\left[\begin{array}{cc}\sec \theta & -\tan \theta \\ -\tan \theta & \sec \theta \end{array}\right]$ is:

• A. $\left[\begin{array}{cc}1& -\sin \theta \\ -\sin \theta & 1\end{array}\right]$
• B. $\left[\begin{array}{cc}\cos \theta & -\sin \theta \\ -\sin \theta & \cos \theta \end{array}\right]$
• C. $\left[\begin{array}{cc}\sec \theta & \tan \theta \\ \tan \theta & \sec \theta \end{array}\right]$
• D. $\left[\begin{array}{cc}\tan \theta & \sec \theta \\ \sec \theta & \tan \theta \end{array}\right]$

Answer 1

Answer: (C)

$\left[\begin{array}{cc}\sec \theta & \tan \theta \\ \tan \theta & \sec \theta \end{array}\right]$

formula for inverse is if $A=\left(\begin{array}{cc}a& b\\ c& d\end{array}\right)$

then $A^{-1}=\frac{1}{ad-bc}\left(\begin{array}{cc}d& -b\\ -c& a\end{array}\right)$

in this $A=\left(\begin{array}{cc}sec\theta & -tan\theta \\ -tan\theta & sec\theta \end{array}\right)$

$A^{-1}=\frac{1}{sec\theta \times sec\theta -(-tan\theta )\times (-tan\theta )}\left(\begin{array}{cc}sec\theta & tan\theta \\ tan\theta & sec\theta \end{array}\right)$

as we know $se{c}^2\theta -ta{n}^2\theta =1$

hence the answer will be $A^{-1}=\left(\begin{array}{cc}sec\theta & tan\theta \\ tan\theta & sec\theta \end{array}\right)$