Show that $\pi =3\arccos(\frac{5}{\sqrt{28}}) + 3\arctan(\frac{\sqrt{3}}{2})$ 
Question:
Show that, $$\pi =3\arccos(\frac{5}{\sqrt{28}}) +
 3\arctan(\frac{\sqrt{3}}{2}) ~~~~~~ (*)$$

My proof method for this question has received mixed responses. Some people say it's fine, others say that it is a verification, instead of a proof.
Proof: $$\pi =3\arccos(\frac{5}{\sqrt{28}}) + 3\arctan(\frac{\sqrt{3}}{2})\iff \frac{\pi}{3} = \arccos(\frac{5}{\sqrt{28}}) + \arctan(\frac{\sqrt{3}}{2}) 
$$$$\iff \frac{\pi}{3} = \arctan(\frac{\sqrt{3}}{5})+\arctan(\frac{\sqrt{3}}{2})$$
As $\arccos(\frac{5}{\sqrt{28}})=\arctan(\frac{\sqrt{3}}{5})$
The plan now is to apply the tangent function to both sides, and show that LHS=RHS using the tangent addition formula to expand it out.
I.e. $$\tan(\frac{\pi}{3}) = \tan\bigg(\arctan(\frac{\sqrt{3}}{5})+\arctan(\frac{\sqrt{3}}{2}\bigg)$$
$$\iff \sqrt{3} = \frac{\frac{\sqrt{3}}{5}+\frac{\sqrt{3}}{2}}{1-\frac{\sqrt{3}}{5} \frac{\sqrt{3}}{2}}$$
and the RHS will reduce down to $\sqrt{3}$. Hence LHS=RHS.
Some things that I've noticed about this method of proof:

*

*It could be used to (incorrectly) prove that $$\frac{\pi}{3}+\pi = \arccos(\frac{5}{\sqrt{28}}) + \arctan(\frac{\sqrt{3}}{2})$$
So because this method of proof can be used to prove things true, that are obviously false, that means it can't be used?

*

*Instead of proving (*), wouldn't this method of proof actually prove that? $$\arccos(\frac{5}{\sqrt{28}})+\arctan(\frac{\sqrt{3}}{2})=\frac{\pi}{3} + \pi k$$
for some $k\in \mathbb{Z}$ which we must find. In this case being when $k=0$.
 A: Using complex numbers:
$$
\arctan(\frac{\sqrt{3}}{5})+\arctan(\frac{\sqrt{3}}{2})
= \arg((5+\sqrt{3}i)(2+\sqrt{3}i))
= \arg(7+7\sqrt{3}i)
= \arctan(\sqrt{3})
$$
A: To show:
$$\text{arctan}\left(\frac{\sqrt{3}}{5}\right) ~+~ \text{arctan}\left(\frac{\sqrt{3}}{2}\right) ~~<~~ \frac{\pi}{2}.$$
In fact this conclusion is immediate by the following analysis.
Let $~\displaystyle \theta ~~=~~ \text{arctan}\left(\frac{\sqrt{3}}{5}\right) ~+~ \text{arctan}\left(\frac{\sqrt{3}}{2}\right).$
Then,
$$\tan(\theta) = 
\frac{\frac{\sqrt{3}}{5} + \frac{\sqrt{3}}{2}}{1 - \left[\frac{\sqrt{3}}{5} \times \frac{\sqrt{3}}{2}\right]}. \tag1 $$
In (1) above, the numerator is clearly positive.
Further, since $~\displaystyle \left[\frac{\sqrt{3}}{5} \times \frac{\sqrt{3}}{2}\right] = \frac{9}{10} < 1$,
the denominator in (1) above is also clearly positive.
Therefore, since $\tan(\theta) > 0$, either $\theta$ is in the 1st quadrant, or $\theta$ is in the 3rd quadrant.
However, by the definition of the arctan function, $\theta$ is the sum of two angles, each of which are between $(0)$ and $(\pi/2)$.  Therefore, $\theta$ can not be in the 3rd quadrant.  Therefore, $\theta$ must be in the first quadrant.

Edit
Thanks to ryang for pointing out a couple of analytical gaps that need to be filled:

*

*By definition, $\text{arccos}\left(\frac{5}{\sqrt{28}}\right)$ is in the 1st quadrant.  Similarly $\text{arctan}\left(\frac{3}{\sqrt{2}}\right)$ is also in the 1st quadrant.  Therefore, $\theta$ is in fact the sum of $2$ angles, each of which are in the first quadrant.


*In general $\tan(\theta) = \tan(\pi/3)$ does not imply that $\theta = (\pi/3)$.  However, if you add the constraint that $\theta$ is in the 1st quadrant, then the implication holds.  Since I deduced that $\theta$ is in the 1st quadrant, the analysis holds.
A: You have$$3\arccos\left(\frac5{\sqrt{28}}\right)\in[0,3],$$since $\frac{\sqrt3}2<\frac5{\sqrt{28}}<1$, and therefore$$3>3\frac\pi6>3\arccos\left(\frac5{\sqrt{28}}\right)>0.$$You also have$$0\leqslant3\arctan\left(\frac{\sqrt3}2\right)<3\arctan\left(\sqrt3\right)=\pi,$$and therefore$$\arccos\left(\frac5{\sqrt{28}}\right)+\arctan\left(\sqrt3\right)\in\left[0,\frac\pi3+1\right].$$What you did shows that$$\tan\left(\arccos\left(\frac5{\sqrt{28}}\right)+\arctan\left(\sqrt3\right)\right)=\tan\left(\frac\pi3\right).$$But the only number in $\left[0,\frac\pi3+1\right]$ whose tangent is $\tan\left(\frac\pi3\right)$ is $\frac\pi3$. So, your proof is indeed correct.
