If A = intlimits_1^{sin theta } {frac{t}{{1 + | vedclass

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(D) First,evaluate $A = \int\limits_1^{\sin 	heta } {\frac{t}{{1 + {t^2}}}} dt$. Let $u = 1 + t^2$,then $du = 2t dt$. So,$A = \frac{1}{2} \ln(1 + t^2

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$2A^3$

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(D) First,evaluate $A = \int\limits_1^{\sin 	heta } {\frac{t}{{1 + {t^2}}}} dt$. Let $u = 1 + t^2$,then $du = 2t dt$. So,$A = \frac{1}{2} \ln(1 + t^2) \Big|_1^{\sin 	heta} = \frac{1}{2} \ln(1 + \sin^2 	heta) - \frac{1}{2} \ln(2) = \frac{1}{2} \ln\left(\frac{1 + \sin^2 	heta}{2}ight)$.Next,evaluate $B = \int\limits_1^{\csc 	heta } {\frac{dt}{{t(1 + t^2)}}}$. Using partial fractions,$\frac{1}{t(1+t^2)} = \frac{1}{t} - \frac{t}{1+t^2}$.$B = \int\limits_1^{\csc 	heta } (\frac{1}{t} - \frac{t}{1+t^2}) dt = [\ln|t| - \frac{1}{2} \ln(1+t^2)]_1^{\csc 	heta} = [\ln(\frac{t}{\sqrt{1+t^2}})]_1^{\csc 	heta}$.$B = \ln(\frac{\csc 	heta}{\sqrt{1+\csc^2 	heta}}) - \ln(\frac{1}{\sqrt{2}}) = \ln(\frac{1}{\sqrt{\sin^2 	heta + 1}}) + \ln(\sqrt{2}) = \ln(\sqrt{\frac{2}{1+\sin^2 	heta}}) = -\frac{1}{2} \ln(\frac{1+\sin^2 	heta}{2})$.Thus,$A = -B$,which implies $A + B = 0$.Substituting $A+B=0$ into the determinant,the term $e^{A+B} = e^0 = 1$.The determinant becomes $\Delta = \left| {\begin{array}{*{20}{c}} A & {{A^2}} & { - B} \ 1 & {{B^2}} & { - 1} \ 1 & {{A^2} + {B^2}} & { - 1} \end{array}} ight|$.Since $A = -B$,we have $A^2 = B^2$ and $-B = A$.$\Delta = \left| {\begin{array}{*{20}{c}} A & {{A^2}} & A \ 1 & {{A^2}} & { - 1} \ 1 & {2{A^2}} & { - 1} \end{array}} ight|$.Subtracting row $2$ from row $3$: $\Delta = \left| {\begin{array}{*{20}{c}} A & {{A^2}} & A \ 1 & {{A^2}} & { - 1} \ 0 & {{A^2}} & 0 \end{array}} ight|$.Expanding along the third row: $\Delta = -A^2 \cdot (-A - A) = -A^2(-2A) = 2A^3$.

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