Find $\arctan(\frac{1}{3})+\arctan(\frac{1}{9})+\arctan(\frac{7}{19})$

Question

Firstly used this formula $$ \begin{align} \arctan(\alpha)+\arctan(\beta) & =\arctan(\frac{1-xy}{x+y}),\quad x\gt0,y\gt0 \\ &=\arctan(\frac{1-\frac{1}{3}\frac{1}{9}}{\frac{1}{3}+\frac{1}{9}}) \\ &=\arctan(2) \end{align}$$ So it is $\arctan(2)+\arctan(\frac{7}{19}).$ Here I don't know what is the next step to solve it completely. A SIDE NOTE: AN EDIT HAS BEEN MADE TO THIS POST, I HAVE FOUND MY MISTAKE! NOW IT IS CLEAR TO ME, THANKS!

Answer 1

The correct formula is: $$\arctan(u)+\arctan(v)=\arctan \left({\frac {u+v}{1-uv}}\right)$$

So: $$\arctan(1/3)+\arctan(1/9)=\arctan \left({\frac {6}{13}}\right)$$ $$\arctan(1/3)+\arctan(1/9)+\arctan(7/19)= \arctan \left({\frac {6/13+7/19}{1-6/13\cdot 7/19}}\right)$$ $$=\arctan(1)=\frac{\pi}{4}$$

Answer 2

Just calculate: $$\tan\left(\arctan\frac{1}{3}+\arctan\frac{1}{9}+\arctan\frac{7}{19}\right)=$$ $$=\frac{\frac{\frac{1}{3}+\frac{1}{9}}{1-\frac{1}{3}\cdot\frac{1}{9}}+\frac{7}{19}}{1-\frac{\frac{1}{3}+\frac{1}{9}}{1-\frac{1}{3}\cdot\frac{1}{9}}\cdot\frac{7}{19}}=1,$$ which gives the answer: $45^{\circ}$.