If $A = \int\limits_1^{\sin \theta } {\frac{t}{{1 + {t^2}}}} dt$ and $B = \int\limits_1^{\cos ec\theta } {\frac{dt}{{t\left( {1 + {t^2}} \right)}}} $ , (where $\theta \in \left( {0,\frac{\pi }{2}} \right))$, then the-value of $\left| {\begin{array}{*{20}{c}}
A&{{A^2}}&{ - B}\\
{{e^{A + B}}}&{{B^2}}&{ - 1}\\
1&{{A^2} + {B^2}}&{ - 1}
\end{array}} \right|$ is
$0$
$A^2$
$A^3$
$2A^3$
Let $m$ and $M$ be respectively the minimum and maximum values of
$\left|\begin{array}{ccc}\cos ^{2} x & 1+\sin ^{2} x & \sin 2 x \\ 1+\cos ^{2} x & \sin ^{2} x & \sin 2 x \\ \cos ^{2} x & \sin ^{2} x & 1+\sin 2 x\end{array}\right|$.
Then the ordered pair $( m , M )$ is equal to
Consider system of equations in $x$ , $y$ and $z$
$12x + by + cz = 0$ ; $ax + 24y + cz = 0$ ; $ax + by + 36z = 0$ .
(where $a$ , $b$ , $c$ are real numbers, $a \ne 12$ , $b \ne 24$ , $c \ne 36$ ).
If system of equation has solution and $z \ne 0$, then value of $\frac{1}{{a - 12}} + \frac{2}{{b - 24}} + \frac{3}{{c - 36}}$ is
If the following system of linear equations
$2 x+y+z=5$
$x-y+z=3$
$x+y+a z=b$
has no solution, then :
Find the area of the triangle whose vertices are $(3,8),(-4,2)$ and $(5,1)$