<s>[INST] Compute: $\arctan{\frac{1}{7}}+\arctan{\frac{3}{4}}.$ 
Compute: $\arctan{\frac{1}{7}}+\arctan{\frac{3}{4}}.$

I want to compute this sum by computing one term at a time. It's clear that 
$$\arctan{\frac{1}{7}}=A \Longleftrightarrow\tan{A}=\frac{1}{7}\Longrightarrow A\in\left(0,\frac{\pi}{2}\right).$$
Drawing a right triangle with sides $1$, $7$ and $5\sqrt{2},$ I get that $$\begin{array}{lcl}
\sin{A} & = & \frac{1}{5\sqrt{2}}\Longleftrightarrow A= \arcsin{\frac{1}{5\sqrt{2}}} \\
\cos{A} & = & \frac{7}{5\sqrt{2}}\Longleftrightarrow A= \arccos{\frac{7}{5\sqrt{2}}} \\
\end{array}$$
But this will not get me standard angles for $A.$
 [/INST] Hint
$$\arctan x +\arctan y
 =\arctan\left(\frac{x+y}{1-xy}\right) $$
See [Additivity of $\arctan(\frac{x+y}{1-xy})$]
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