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

  • A

    $0$

  • B

    $A^2$

  • C

    $A^3$

  • D

    $2A^3$

Similar Questions

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

 

  • [JEE MAIN 2020]

Let $A =$ $\left[ {\begin{array}{*{20}{c}}{1 + {x^2} - {y^2} - {z^2}}&{2(xy + z)}&{2(zx - y)}\\{2(xy - z)}&{1 + {y^2} - {z^2} - {x^2}}&{2(yz + x)}\\{2(zx + y)}&{2(yz - x)}&{1 + {z^2} - {x^2} - {y^2}}\end{array}} \right]$  then det. $A$ 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 :

  • [JEE MAIN 2021]

Find the area of the triangle whose vertices are $(3,8),(-4,2)$ and $(5,1)$