Given $\tan a = \frac{1}{7}$ and $\sin b = \frac{1}{\sqrt{10}}$, show $a+2b = \frac{\pi}{4}$. 
Given
  $$\tan a = \frac{1}{7} \qquad\text{and}\qquad \sin b = \frac{1}{\sqrt{10}}$$
  $$a,b \in (0,\frac{\pi}{2})$$
  Show that $$a+2b=\frac{\pi}{4}$$

Does exist any faster method of proving that, other than expanding $\sin{(a+2b)}$?
Thank You!

$\sin b = \dfrac{1}{\sqrt{10}}$, $b\in(0,\pi/2)$ $\;\Rightarrow\;$ $b\in(0,\pi/6)$, i.e. $2b\in (0,\pi/3)$, since $1/\sqrt{10}<1/2$, and
$$\sin 2b = 2\sin b \cos b = 2 \dfrac{1}{\sqrt{10}} \dfrac{3}{\sqrt{10}} = \dfrac{3}{5}.$$
Then
$$
\tan 2b = \dfrac{\sin 2b}{\cos 2b} = \dfrac{3/5}{4/5} = \dfrac{3}{4}.
$$
Now use formula
$$
\tan(\alpha+\beta) = \dfrac{\tan \alpha + \tan \beta}{1-\tan \alpha \tan \beta},
$$
$$
\tan(a+2b) = \dfrac{\tan a + \tan 2b}{1-\tan a \tan 2b} = \dfrac{\frac{1}{7} + \frac{3}{4}}{1 - \frac{1}{7}\cdot\frac{3}{4}} = \dfrac{4+21}{28-3}=\dfrac{25}{25}=1.
$$
Then , in general, $a+2b = \pi/4+\pi k$, $k \in \mathbb{Z}$. 
But since $a$ and $2b$ are acute, then $a+2b=\pi/4$. 
