If A = ∫limits₁^{sin θ } {frac{t}{{1 + {t^2}}}} dt and B = ∫limits₁^{cos ecθ }

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3 and 4 .Determinants and Matrices

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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

$2A^3$

Solution

$A = \int\limits_1^{\sin \theta } {\frac{1}{{1 + {t^2}}}dt} $

put $t = \frac{1}{z}$

$A = \int\limits_1^{\cos ec\theta } { – \frac{{dz}}{{z\left( {{z^2} + 1} \right)}}} $

$A = – B \Rightarrow A + B = 0$

Standard 12

Mathematics

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