Question

If ${\Delta }_1=\left|\begin{array}{ccc}x& \sin \theta & \cos \theta \\ -\sin \theta & -x& 1\\ \cos \theta & 1& x\end{array}\right|$ and ${\Delta }_2=\left|\begin{array}{ccc}x& \sin 2\theta & \cos 2\theta \\ -\sin 2\theta & -x& 1\\ \cos 2\theta & 1& x\end{array}\right|$ , $x\ne 0$ ; then for all $\theta \in (0,\frac{\pi }{2})$ :

• A. ${\Delta }_1-{\Delta }_2=-2x^3$
• B. ${\Delta }_1+{\Delta }_2=-2x^3$
• C. ${\Delta }_1+{\Delta }_2=-2(x^3+x-1)$
• D. ${\Delta }_1-{\Delta }_2=x(\cos 2\theta -\cos 4\theta )$

Answer 1

The correct option is B ${\Delta }_1+{\Delta }_2=-2x^3$

On expanding the determinant , we have
${\Delta }_1=x(-x^2-1)-\sin \theta (-x\sin \theta -\cos \theta )+\cos \theta (-\sin \theta +x\cos \theta )=-x^3+x({\sin }^2\theta +{\cos }^2\theta )-x=-x^3$

Similarly, ${\Delta }_2=-x^3$
${\Delta }_1+{\Delta }_2=-2x^3$