Question

Que is on pic.i can't able to type in text.

Answer 1

It's not difficult to compute the values of
`A` and
`B` directly:


`A=\displaystyle\int_1^(\sin\theta)(\mathrm dt)/(1+t^2)=\tan^(-1)t\bigg|_(t=1)^(t=\sin\theta)`

`A=\tan^(-1)(\sin\theta)-\frac\pi4`


`B=\displaystyle\int_1^(\csc\theta)(\mathrm dt)/(t(1+t^2))=\int_1^(\csc\theta)\left(\frac1t-\frac t{1+t^2}\right)\,\mathrm dt`

`B=\left(\ln|t|-\frac12\ln|1+t^2|\right)\bigg|_(t=1)^(t=\csc\theta)`

`B=\ln\left|(\csc\theta)/(√(1+\csc^2\theta))\right|+\frac12\ln2`

Let's assume
`0<\theta<\pi`, so that
`|\csc\theta|=\csc\theta`.

Now,


`\Delta=\begin{vmatrix}A&A^2&B\\e^(A+B)&B^2&-1\\1&A^2+B^2&-1\end{vmatrix}`

`\Delta=A\begin{vmatrix}B^2&-1\\A^2+B^2&-1\end{vmatrix}-e^(A+B)\begin{vmatrix}A^2&B\\A^2+B^2&-1\end{vmatrix}+\begin{vmatrix}A^2&B\\B^2&-1\end{vmatrix}`

`\Delta=A(-B^2+A^2+B^2)-e^(A+B)(-A^2-A^2B-B^3)+(-A^2-B^3)`

`\Delta=A^3-A^2-B^3+e^(A+B)(A^2+A^2B+B^3)`

There doesn't seem to be anything interesting about this result... But all that's left to do is plug in
`A` and
`B`.