<s>[INST] 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!
 [/INST] 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.
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