With usual notations, in $\triangle ABC$, $(b - c)^2 \cos^2 \dfrac{A}{2} + (b + c)^2 \sin^2 \dfrac{A}{2} =$

Let $A = \begin{bmatrix} -5 & -3 \\ 2 & 1 \end{bmatrix}$. The Row transformation $R_1 \rightarrow R_1 + 3R_2$ will transform matrix A into

If $A = \begin{bmatrix} 2 & 0 & -1 \\ 5 & 1 & 0 \\ 0 & 1 & 3 \end{bmatrix}_{3 \times 3}$, and $A^{-1} = \begin{bmatrix} \gamma & -1 & 1 \\ \alpha & 6 & -5 \\ \beta & -2 & 2 \end{bmatrix}_{3 \times 3}$, then $|\alpha \cdot \beta \cdot \gamma| = $ _____ (where $|\cdot|$ denotes the absolute value)

If $y = \tan^{-1}\left\{\dfrac{\sqrt{1 + x^2} - \sqrt{1 - x^2}}{\sqrt{1 + x^2} + \sqrt{1 - x^2}}\right\}$, where $|x| < 1$, then $\dfrac{dy}{dx}$ is equal to