If A = int_{1}^{sin theta} frac{t}{1+t^2} dt | vedclass

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(A) First,evaluate $A$: $A = \int_{1}^{\sin 	heta} \frac{t}{1+t^2} dt = \frac{1}{2} [\ln(1+t^2)]_{1}^{\sin 	heta} = \frac{1}{2} \ln(1+\sin^2 	heta)

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(A) First,evaluate $A$: $A = \int_{1}^{\sin 	heta} \frac{t}{1+t^2} dt = \frac{1}{2} [\ln(1+t^2)]_{1}^{\sin 	heta} = \frac{1}{2} \ln(1+\sin^2 	heta) - \frac{1}{2} \ln(2) = \frac{1}{2} \ln(\frac{1+\sin^2 	heta}{2})$.Next,evaluate $B$: $B = \int_{1}^{\operatorname{cosec} 	heta} \frac{1}{t(1+t^2)} dt$. Using partial fractions,$\frac{1}{t(1+t^2)} = \frac{1}{t} - \frac{t}{1+t^2}$.So,$B = [\ln|t| - \frac{1}{2} \ln(1+t^2)]_{1}^{\operatorname{cosec} 	heta} = [\ln(\frac{t}{\sqrt{1+t^2}})]_{1}^{\operatorname{cosec} 	heta}$.Since $\frac{t}{\sqrt{1+t^2}} = \frac{\operatorname{cosec} 	heta}{\sqrt{1+\operatorname{cosec}^2 	heta}} = \frac{1}{\sqrt{\sin^2 	heta + 1}}$,we have $B = \ln(\frac{1}{\sqrt{1+\sin^2 	heta}}) - \ln(\frac{1}{\sqrt{2}}) = \ln(\sqrt{\frac{2}{1+\sin^2 	heta}}) = -\frac{1}{2} \ln(\frac{1+\sin^2 	heta}{2}) = -A$.Thus,$A+B = 0$,which implies $e^{A+B} = e^0 = 1$.The determinant becomes $\left| \begin{array}{ccc} A & A^2 & -A \ 1 & (-A)^2 & -1 \ 1 & A^2+(-A)^2 & -1 \end{array} ight| = \left| \begin{array}{ccc} A & A^2 & -A \ 1 & A^2 & -1 \ 1 & 2A^2 & -1 \end{array} ight|$.Since the first and third columns are proportional (column $3$ is $-1$ times column $1$),the determinant is $0$.

If $A = \int_{1}^{\sin \theta} \frac{t}{1+t^2} dt$ and $B = \int_{1}^{\operatorname{cosec} \theta} \frac{1}{t(1+t^2)} dt$,then the value of $\left| \begin{array}{ccc} A & A^2 & B \\ e^{A+B} & B^2 & -1 \\ 1 & A^2+B^2 & -1 \end{array} \right| = $

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