Simplifying arctan expressions such as $\arctan 8 + \arctan 2 + \arctan\frac23$

Question

How does one simplify an arctan expression? Are there identities that exist for arctan to bring them together?

For example, an exercise asked for $$\arctan 8 + \arctan 2 + \arctan\frac23$$ in terms of $\pi$, but no calculators are allowed.

I've never seen or heard of arctan identities.

Answer 1

Hint:

$$\tan(\arctan 8+\arctan2)=\frac{8+2}{1-8\cdot2}=-\frac{2}{3}$$ so that

$$\arctan 8+\arctan2=k\pi-\arctan\frac23.$$

Answer 2

If you can use complex numbers, then $$ \arctan 8 + \arctan 2 + \arctan\frac23 = \arg ((1+8i)(1+2i)(3+2i)) = \arg(-65) = \pi $$

Answer 3

Using Are $\mathrm{arccot}(x)$ and $\arctan(1/x)$ the same function?,

for $x>0$ $\arctan x=$arccot$\dfrac1x=\dfrac\pi2-\arctan\dfrac1x$

$$\implies\arctan8+\arctan2=\pi-\arctan\dfrac12-\arctan\dfrac18$$

Using Inverse trigonometric function identity doubt: $\tan^{-1}x+\tan^{-1}y =-\pi+\tan^{-1}\left(\frac{x+y}{1-xy}\right)$, when $x<0$, $y<0$, and $xy>1$,

$$\implies\arctan\dfrac12+\arctan\dfrac18=\arctan\dfrac{\dfrac12+\dfrac18}{1-\dfrac12\cdot\dfrac18}=\arctan\dfrac{10}{16-1}=\arctan\dfrac23$$

Hope you can take it from here!