Solving for trigonometric values Find $a+2b$ if $\tan a = 1/7$ and $\sin b=\frac{1}{\sqrt{10}}$. I had tried to solve it by trigonometric ratios but i could not. Please solve it by a method of class10 standards.
 A: Hint:
$\tan a = \frac{1}{7}$
$\tan b = \frac{1}{3}$
$\tan(a+2b) = \frac{\tan a + \tan 2b}{1-\tan a \tan 2b}$
$\tan 2b = \frac{2tanb}{1-tan^2 b}$
You get $\tan 2b = \frac{6}{8}$
$\tan (a + 2b) = 1$
$a+2b = \arctan{1} = \frac{\pi}{4}$
A: $$\tan(a)=\frac{1}{7}\Longleftrightarrow a=\pi n+\tan^{-1}\left(\frac{1}{7}\right)$$
$$\sin(b)=\frac{1}{\sqrt{10}}\Longleftrightarrow b=\{_{2\pi n+\pi-\sin^{-1}\left(\frac{1}{\sqrt{10}}\right)}^{2\pi n+\sin^{-1}\left(\frac{1}{\sqrt{10}}\right)}$$
With $n\in \mathbb{Z}$

$$a+2b\Longrightarrow$$
1)
$$\left(\pi n+\tan^{-1}\left(\frac{1}{7}\right)\right)+2\left(2\pi n+\pi-\sin^{-1}\left(\frac{1}{\sqrt{10}}\right)\right)=$$
$$5\pi n+2\pi-2\sin^{-1}\left(\frac{1}{\sqrt{10}}\right)+\tan^{-1}\left(\frac{1}{7}\right)$$
2)
$$\left(\pi n+\tan^{-1}\left(\frac{1}{7}\right)\right)+2\left(2\pi n+\sin^{-1}\left(\frac{1}{\sqrt{10}}\right)\right)=$$
$$5\pi n+2\sin^{-1}\left(\frac{1}{\sqrt{10}}\right)+\tan^{-1}\left(\frac{1}{7}\right)$$
A: $$\cos(b)=\pm\sqrt{1-\sin^2(b)}=\pm\frac3{\sqrt{10}}.$$
$$\tan(2b)=\frac{\sin(2b)}{\cos(2b)}=\frac{2\sin(b)\cos(b)}{1-2\sin^2(b)}=\pm\frac34.$$
$$\tan(a+2b)=\frac{\tan(a)+\tan(2b)}{1-\tan(a)\tan(2b)}=1\text{ or }-\frac{17}{31}.$$
