e sec h - 1 1 2 + tan h - 1 1 2 + sin h - 1 1 2 =

We know that,

$\sec h^{-1}x=\log \left(\frac{1+\sqrt{1-x^2}}{x}\right)$

$\tan h^{-1}x=\frac{1}{2}\log \left(\frac{1+x}{1-x}\right)$

$\sin h^{-1}x=\log \left(x+\sqrt{x^2+1}\right)$

Substitute the values in the given equation and simplify.

$e^{\log \left(\frac{1+\sqrt{1-{\displaystyle \frac{1}{4}}}}{{\displaystyle \frac{1}{2}}}\right)·\frac{1}{2}\log \left(\frac{1+{\displaystyle \frac{1}{2}}}{1-\frac{1}{2}}\right)+\log \left(\frac{1}{2}+\sqrt{\frac{1}{4}+1}\right)}=\left(2+\sqrt{3}\right)\left(\sqrt{3}\right)\left(\frac{\sqrt{5}+1}{2}\right)$

$=\frac{\left(2\sqrt{3}+3\right)\left(\sqrt{5}+1\right)}{2}$

$=\frac{2\sqrt{15}+2\sqrt{3}+3\sqrt{5}+3}{2}$

$=\frac{3+3\sqrt{5}+2\sqrt{3}+2\sqrt{15}}{2}$