Given,

$\tan hx=\frac{1}{2}$

Then $\sin hx=\frac{1}{\sqrt{3}}$ and $\cos hx=\frac{2}{\sqrt{3}}$

Now the value of $\sin h2x-\sec h2x=2\sin hx\cos hx-\frac{1}{{\cos }^{2}hx+{\sin }^{2}hx}$

{using the formula $\sin h\left(x+y\right)=\sin hx\cos hy+\cos hx\sin hy$ and $\cos h\left(x+y\right)=\cos hx\cos hy+\sin hx\sin hy$}

$\Rightarrow \sin h2x-\sec h2x=2\times \frac{1}{\sqrt{3}}\times \frac{2}{\sqrt{3}}-\frac{1}{{\left({\displaystyle \frac{{\displaystyle 2}}{{\displaystyle \sqrt{3}}}}\right)}^2+{\left({\displaystyle \frac{1}{\sqrt{3}}}\right)}^2}$

$\Rightarrow \sin h2x-\sec h2x=\frac{4}{3}-\frac{3}{5}$

$\Rightarrow \sin h2x-\sec h2x=\frac{11}{15}$