Q:

Sum of all possible angles.

If $$\tan\left(\frac{\pi}{12}-x\right) , \tan\left(\frac{\pi}{12}\right) , \tan \left(\frac{\pi}{12} +x\right)$$ in order are three consecutive terms of a GP then what is sum of all possible values of $x$.
I am not getting any start, can anybody provide me a hint?

A:

Use :$$\Big(\tan\frac{\pi}{2}\Big)^2=\tan\Big(\frac{\pi}{12}-x\Big)\tan\Big(\frac{\pi}{12}+x\Big) \tag1$$ (Condition for G.P)
And :
$$\tan (A\pm B) = \frac{\tan A \pm \tan B}{1\mp \tan A\tan B}$$
Let $\tan \dfrac{\pi}{12} =c$
Put in $(1)$ 
$$c^2=\frac {c+\tan x}{1-c\tan x} \cdot \frac{c-\tan x}{1+c\tan x} \implies c^2-c^4\tan^2 x =c^2-\tan^2 x \implies \tan x =0$$
Sum of solutions in a bounded interval will be sum all angles in the form $k\pi ~;~ k \in \mathbb Z$

