The negation of $(p \wedge q) \rightarrow ((p \vee r) \rightarrow\, \sim q)$ is equivalent to ...

The statement pattern $[(p \wedge q) \rightarrow (\sim p \vee r)] \vee [(\sim p \vee r) \rightarrow (p \wedge q)]$ is

In $\triangle ABC$, with usual notation, if cot A, cot B, cot C are in arithmetic progression, then

If $A = \begin{bmatrix} 1 & -\tan\dfrac{\theta}{2} \\ \tan\dfrac{\theta}{2} & 1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & \tan\dfrac{\theta}{2} \\ -\tan\dfrac{\theta}{2} & 1 \end{bmatrix}$ then $A^{-1}B$ is equal to