<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] 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 $xy>1$,
$a+2b=\arctan\dfrac17+\arctan\dfrac34=\arctan\dfrac{\dfrac17+\dfrac34}{1-\dfrac17\cdot\dfrac34}=\arctan1=\dfrac\pi4$
</s>