[tex]\sin(a-b)=\sin a \cos b-\cos a \sin b[/tex]. If we let [tex]a=\cos^{-1} \left(\frac{5}{6} \right)[/tex] and [tex]b=\tan^{-1} \left(\frac{1}{2} \right)[/tex], then the given expression is equal to:
[tex]\sin \left(\cos^{-1} \left(\frac{5}{6}} \right) \right) \cos \left(\tan^{-1} \left(\frac{1}{2} \right) \right)-\cos\left(\cos^{-1} \left(\frac{5}{6} \right) \right) \sin \left( \tan^{-1} \left(\frac{1}{2} \right) \right)[/tex]
Using the Pythagorean identities [tex]\sin^{2} x+\cos^{2} x=1[/tex] and [tex]\tan^{2} x+1=\sec^{2} x[/tex],
[tex]1) \sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)+\cos^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=1\\\sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)+\frac{25}{36}=1\\\sin^{2} \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=\frac{11}{36}\sin \left(\cos^{-1} \left(\frac{5}{6} \right) \right)=\frac{\sqrt{11}}{6}[/tex]
[tex]2) \tan^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)+1=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\frac{1}{4}+1=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\frac{5}{4}=\sec^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)\\\sec \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{\sqrt{5}}{2}\\\implies \cos \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}[/tex]
[tex]\cos^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)+\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=1\\\frac{4}{5}+\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=1\\\sin^{2} \left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{1}{5}\\\left(\tan^{-1} \left(\frac{1}{2} \right) \right)=\frac{1}{\sqrt{5}}=\frac{\sqrt{5}}{5}[/tex]
This means we can write the original expression as:
[tex]\left(\frac{\sqrt{11}}{6} \right) \left(\frac{2\sqrt{5}}{5} \right)-\left(\frac{5}{6} \right) \left(\frac{\sqrt{5}}{5} \right)\\=\frac{2\sqrt{11}\sqrt{5}}{30}-\frac{5\sqrt{5}}{30}\\=\boxed{\frac{\sqrt{5}(2\sqrt{11}-5)}{30}}[/tex]
