$\left|\begin{array}{ccc}\cos 3\pi & \sin 5\pi & \tan 7\pi \\ \sqrt{3} & 1 & 0 \\ \sqrt{5} & 0 & 1\end{array}\right| = $ . . . . . . .

If $A$,$B$ and $C$ are the angles of a triangle,then the value of the determinant $\left| \begin{array}{ccc} -1 + \cos B & \cos C + \cos B & \cos B \\ \cos C + \cos A & -1 + \cos A & \cos A \\ -1 + \cos B & -1 + \cos A & -1 \end{array} \right|$ is:

Let $A = \begin{bmatrix} 1 & \sin \theta & 1 \\ -\sin \theta & 1 & \sin \theta \\ -1 & -\sin \theta & 1 \end{bmatrix}$,where $0 \le \theta < 2\pi$,then:

If $a, b, c$ are respectively the $5^{\text{th}}, 8^{\text{th}}, 13^{\text{th}}$ terms of an arithmetic progression,then $\left|\begin{array}{ccc}a & 5 & 1 \\ b & 8 & 1 \\ c & 13 & 1\end{array}\right|=$

If the area of a triangle is $35$ $sq$ $units$ with vertices $(2,-6), (5,4)$ and $(k, 4)$,then $k$ is:

If $f(x) = \left| \begin{array}{ccc} 1 & 6+x & 36+x^2 \\ 0 & x-3 & 3x^2-27 \\ 0 & 2x-4 & 8x^2-32 \end{array} \right|$,then $\lim_{x \rightarrow 1} \frac{f(x)}{f(-x)} = $