Question

For all values of $\Theta \in (0,\frac{\pi }{2})$ the determinants of the matrix $\left[\begin{array}{ccc}-2& \tan \theta +{\sec }^2\theta & 3\\ -\sin \theta & \cos \theta & \sin \theta \\ -3& 4& 3\end{array}\right]$ always lies in the interval:

• A. $(-4,3)$
• B. $(1,2)$
• C. $(4,3)$
• D. $(4,5)$

Answer 1

The correct option is A $(-4,3)$

We have,

$\begin{array}{c}\left|\begin{array}{ccc}-2& \tan \theta +{\sec }^2\theta & 3\\ -\sin \theta & \cos \theta & \sin \theta \\ -3& 4& 3\end{array}\right|=-2\left(3\cos \theta -4\sin \theta \right)-\left({\sec }^2\theta +\tan \theta \right)\left(0\right)+3\left(-4\sin \theta +3\cos \theta \right)\\ =-6\cos \theta +8\sin \theta -12\sin \theta +9\cos \theta \\ =-4\sin \theta +3\cos \theta \\ f\left(\theta \right)=3\cos \theta -4\sin \theta \\ whichalwaysliesbetween\left(-4,3\right)\end{array}$

Hence, option $A$ is the correct answer