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!
 A: $\sin b = \dfrac{1}{\sqrt{10}}$, $b\in(0,\pi/2)$ $\;\Rightarrow\;$ $b\in(0,\pi/6)$, 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 $a+2b=\pi/4$. 
A: Just for fun...  Adding a picture for the first time, so bear with me.

$\sin b = 1/\sqrt{10} \implies \cos b = 3/\sqrt{10}$ and $\sin 2b = 3/5$. 
Now consider an isosceles right triangle $ABC$ as in the picture, with $AB = AC = 4$. Pick $D$ on $AC$ so that $AD = 3$, and $DE$ perpendicular to $BC$. $BD = 5$ by the Pythagorean theorem.  
$\sin \angle DBA = 3/5$, $\angle ABC = \pi/4$, so it suffices to show $a = \angle DBE$. Now the area of $\triangle ABC$ is $8$ and that of $\triangle ABD$ is $6$, so the area of $\triangle DBC$ is $2$, and $BC = 4\sqrt 2$. From the formula for the area, $DE = 1/\sqrt 2$ , and $EB = 7/\sqrt2$ (by the Pythagorean theorem), $\tan \angle DBE = 1/7$, and the conclusion follows.
ADDED: $DE = 1/\sqrt2$ follows, more directly, from the fact that $\triangle ECD$ is isosceles and right-angled - no need for all the nonsense about areas.
A: So, $\cos b=+\sqrt{1-\left(\dfrac1{\sqrt{10}}\right)^2}=\dfrac3{\sqrt{10}}$
$\implies\tan b=\dfrac{\dfrac1{\sqrt{10}}}{\dfrac3{\sqrt{10}}}\implies b=\arctan \dfrac13$
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 $xy>1$,
$$2b=\arctan\dfrac{2\cdot\dfrac13}{1-\left(\dfrac13\right)^2}=\arctan\dfrac34$$
Again as $a=\arctan\dfrac17$
Again 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 $xy>1$,
$a+2b=\arctan\dfrac17+\arctan\dfrac34=\arctan\dfrac{\dfrac17+\dfrac34}{1-\dfrac17\cdot\dfrac34}=\arctan1=\dfrac\pi4$
