Prove that :
$$\frac{1}{2} \tan^{-1}(2)+\frac{1}{3} \tan^{-1}(3)-2\tan^{-1}\left(\frac{1}{2}\right)-3\tan^{-1}\left(\frac{1}{3}\right)=-\frac{5 \pi}{8}+\frac{5}{6} \tan^{-1}(3)$$
Using the identities $\tan^{-1}(x)=\cot^{-1} (1/x)$ for $x>0$ and $\cot^{-1}(x)+\tan^{-1}(x)=\pi/2$ I have got $\frac{3}{2} \tan^{-1}(2)+\frac{10}{3} \tan^{-1}(3)- \frac{5 \pi}{2}$ but I am not able to get final result. Could someone help me with it?
Firstly, $$\frac 12\arctan2-2\arctan\frac 12=\frac 12\left(\frac{\pi}{2}-\arctan \frac 12\right)-2\arctan\frac 12=\frac{\pi}{4}-\frac 52\arctan \frac 12$$
Secondly, you should be able to show in a similar way that $$\frac 13\arctan3-3\arctan \frac 13-\frac 56\arctan 3=-\frac{\pi}{4}-\frac 52\arctan\frac 13$$
It then remains to obtain the result by using $$\arctan a+\arctan b=\arctan\left(\frac{a+b}{1-ab}\right)$$