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LammettHash
It's not difficult to compute the values of [tex]A[/tex] and [tex]B[/tex] directly:

[tex]A=\displaystyle\int_1^{\sin\theta}\frac{\mathrm dt}{1+t^2}=\tan^{-1}t\bigg|_{t=1}^{t=\sin\theta}[/tex]
[tex]A=\tan^{-1}(\sin\theta)-\dfrac\pi4[/tex]

[tex]B=\displaystyle\int_1^{\csc\theta}\frac{\mathrm dt}{t(1+t^2)}=\int_1^{\csc\theta}\left(\frac1t-\frac t{1+t^2}\right)\,\mathrm dt[/tex]
[tex]B=\left(\ln|t|-\dfrac12\ln|1+t^2|\right)\bigg|_{t=1}^{t=\csc\theta}[/tex]
[tex]B=\ln\left|\dfrac{\csc\theta}{\sqrt{1+\csc^2\theta}}\right|+\dfrac12\ln2[/tex]

Let's assume [tex]0<\theta<\pi[/tex], so that [tex]|\csc\theta|=\csc\theta[/tex].

Now,

[tex]\Delta=\begin{vmatrix}A&A^2&B\\e^{A+B}&B^2&-1\\1&A^2+B^2&-1\end{vmatrix}[/tex]
[tex]\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}[/tex]
[tex]\Delta=A(-B^2+A^2+B^2)-e^{A+B}(-A^2-A^2B-B^3)+(-A^2-B^3)[/tex]
[tex]\Delta=A^3-A^2-B^3+e^{A+B}(A^2+A^2B+B^3)[/tex]

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

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