If \(A=\displaystyle \int_{1}^{sin\theta}\frac{t}{1+t^{2}}dt\) and

\(B=\displaystyle \int_{1}^{cosec\theta}\frac{1}{t(1+t^{2})}dt\), then the value of determinant \(\begin{vmatrix}

A & A^{2}& B\\

e^{A+B}& B^{2} &-1 \\

1& A^{2}+B^{2} & -1

\end{vmatrix}\) is

(A) \(\sin\theta\)

(B) \(cosec \theta\)

(C) \(0\)

(D) \(1\)

\(B=\displaystyle \int_{1}^{cosec\theta}\frac{1}{t(1+t^{2})}dt\), then the value of determinant \(\begin{vmatrix}

A & A^{2}& B\\

e^{A+B}& B^{2} &-1 \\

1& A^{2}+B^{2} & -1

\end{vmatrix}\) is

(A) \(\sin\theta\)

(B) \(cosec \theta\)

(C) \(0\)

(D) \(1\)

Answer: C

\(A=\displaystyle \int _{ 1 }^{ sin\theta } \frac { t }{ 1+t^{ 2 } } dt\)

\(\displaystyle B=\int _{ 1 }^{ cosec\theta } \frac { 1 }{ t(1+t^{ 2 }) } dt\)

Put \(\displaystyle z=\frac { 1 }{ t } \quad \)

\(\displaystyle dz=-\frac { 1 }{ { t }^{ 2 } } dt\)

\(\displaystyle B=\int _{ 1 }^{ \sin { \theta } } \frac { -z }{ (z^{ 2 }+1) } dz\)

\(\displaystyle =\int _{ 1 }^{ \sin { \theta } } \frac { -t }{ (t^{ 2 }+1) } dt\)

\(\Rightarrow B=-A\)

Now, \(\begin{vmatrix} A & A^{ 2 } & B \\ e^{ A+B } & B^{ 2 } & -1 \\ 1 & A^{ 2 }+B^{ 2 } & -1 \end{vmatrix}\)

\(=\begin{vmatrix} A & A^{ 2 } & -A \\ e^{ 0 } & A^{ 2 } & -1 \\ 1 & 2A^{ 2 } & -1 \end{vmatrix}\)

\(=-\begin{vmatrix} A & A^{ 2 } & A \\ 1 & A^{ 2 } & 1 \\ 1 & 2A^{ 2 } & 1 \end{vmatrix}\)

\(=0\) (On expanding the determinant)

\(A=\displaystyle \int _{ 1 }^{ sin\theta } \frac { t }{ 1+t^{ 2 } } dt\)

\(\displaystyle B=\int _{ 1 }^{ cosec\theta } \frac { 1 }{ t(1+t^{ 2 }) } dt\)

Put \(\displaystyle z=\frac { 1 }{ t } \quad \)

\(\displaystyle dz=-\frac { 1 }{ { t }^{ 2 } } dt\)

\(\displaystyle B=\int _{ 1 }^{ \sin { \theta } } \frac { -z }{ (z^{ 2 }+1) } dz\)

\(\displaystyle =\int _{ 1 }^{ \sin { \theta } } \frac { -t }{ (t^{ 2 }+1) } dt\)

\(\Rightarrow B=-A\)

Now, \(\begin{vmatrix} A & A^{ 2 } & B \\ e^{ A+B } & B^{ 2 } & -1 \\ 1 & A^{ 2 }+B^{ 2 } & -1 \end{vmatrix}\)

\(=\begin{vmatrix} A & A^{ 2 } & -A \\ e^{ 0 } & A^{ 2 } & -1 \\ 1 & 2A^{ 2 } & -1 \end{vmatrix}\)

\(=-\begin{vmatrix} A & A^{ 2 } & A \\ 1 & A^{ 2 } & 1 \\ 1 & 2A^{ 2 } & 1 \end{vmatrix}\)

\(=0\) (On expanding the determinant)

To Keep Reading This Answer, Download the App

4.6

Review from Google Play

To Keep Reading This Answer, Download the App

4.6

Review from Google Play

Correct39

Incorrect0

Still Have Question?

Load More

More Solution Recommended For You