If $\sin ^{-1}\left(\frac{12}{x}\right)+\sin ^{-1}\left(\frac{5}{x}\right)=\frac{\pi}{2}$, then $x=$

$$ \operatorname{cosec}^{-1}\left[\left(\frac{\tan ^2\left(\frac{\alpha-\pi}{4}\right)-1}{\tan ^2\left(\frac{\alpha-\pi}{4}\right)+1}+\cos \frac{\alpha}{2} \cdot \cot 5 \alpha\right) \sec \frac{11 \alpha}{2}\right] $$

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