The value of $\theta$ lying between $0$ and $\pi / 2$ and satisfying the equation $\left| \begin{array}{ccc} 1 + \sin^2 \theta & \cos^2 \theta & 4 \sin 4 \theta \\ \sin^2 \theta & 1 + \cos^2 \theta & 4 \sin 4 \theta \\ \sin^2 \theta & \cos^2 \theta & 1 + 4 \sin 4 \theta \end{array} \right| = 0$ is:

If $a > 0$ and the discriminant of $ax^2 + 2bx + c$ is negative,then $\left| \begin{array}{ccc} a & b & ax + b \\ b & c & bx + c \\ ax + b & bx + c & 0 \end{array} \right|$ is

If a $2^{nd}$ order determinant with elements $0$ or $1$ is chosen from the set of all such determinants,find the probability that the determinant chosen is non-zero.

Let $\omega$ be a complex number such that $2\omega + 1 = z$ where $z = \sqrt{-3}$. If $\left| \begin{array}{ccc} 1 & 1 & 1 \\ 1 & -\omega^2 - 1 & \omega^2 \\ 1 & \omega^2 & \omega^7 \end{array} \right| = 3k$,then $k$ is equal to:

Let $A = \begin{bmatrix} 1 & -1 \\ 2 & \alpha \end{bmatrix}$ and $B = \begin{bmatrix} \beta & 1 \\ 1 & 0 \end{bmatrix}$,where $\alpha, \beta \in \mathbb{R}$. Let $\alpha_{1}$ be the value of $\alpha$ which satisfies $(A + B)^{2} = A^{2} + \begin{bmatrix} 2 & 2 \\ 2 & 2 \end{bmatrix}$ and $\alpha_{2}$ be the value of $\alpha$ which satisfies $(A + B)^{2} = B^{2}$. Then $|\alpha_{1} - \alpha_{2}|$ is equal to:

If $ A = \frac{1}{\pi} \begin{bmatrix} \sin^{-1}(\pi x) & \tan^{-1}(\frac{x}{\pi}) \\ \tan^{-1}(\frac{x}{\pi}) & \cot^{-1}(\pi x) \end{bmatrix} $ and $ B = \begin{bmatrix} -\cos^{-1}(\pi x) & \tan^{-1}(\frac{x}{\pi}) \\ \sin^{-1}(\frac{x}{\pi}) & -\tan^{-1}(\pi x) \end{bmatrix} $,then $ A - B $ is