If $\tan ^{-1} \frac{1}{5}+\frac{1}{2} \sec ^{-1} x+\tan ^{-1} \frac{1}{8}=\frac{\pi}{8}$, then $x^2=$

Assertion $(\mathrm{A}) \operatorname{cosech}^{-1}(3)=\log \left(\frac{1+\sqrt{10}}{3}\right)$

Reason (R) $e^{\operatorname{cosech}^{-1} x}$ is a root of the quadratic equation $x p^2-2 p-x=0$

The correct option among the following is