Question

For real numbers $p,q,r$ and $s,$ if $({\cos }^{-1}p+{\cos }^{-1}q)({\cos }^{-1}r+{\cos }^{-1}s)=4{\pi }^2$ and $\Delta =\left|\begin{array}{cc}{p}^{n_1}& q^{n_2}\\ r^{n_3}& s^{n_4}\end{array}\right|$ where $n_1,n_2,n_3,n_4\in N,$ then the absolute difference between the maximum and the minimum values of $\Delta$ is equal to

Answer 1

$({\cos }^{-1}p+{\cos }^{-1}q)\cdot ({\cos }^{-1}r+{\cos }^{-1}s)=4{\pi }^2$
We know that $0\le {\cos }^{-1}x\le \pi$
So, this is true only if $p=q=r=s=-1$
${\Delta }_{max}=\left|\begin{array}{cc}1& -1\\ -1& 1\end{array}\right|=2{\Delta }_{min}=\left|\begin{array}{cc}-1& 1\\ 1& 1\end{array}\right|=-2|{\Delta }_{max}-{\Delta }_{min}|=4$