Find value of x, where $\frac{3+\cot 80^{\circ} \cot 20^{\circ}}{\cot 80^{\circ}+\cot 20^{\circ}}=\cot x^{\circ}$ $$\frac{3+\cot 80^{\circ}\cot 20^{\circ}}{\cot 80^{\circ}+\cot 20^{\circ}}=\cot x^{\circ}$$ Then find $x$
My Try:
Using $\cot 80=\tan 10$ and $\cot 20=\frac{1}{\tan 20}$
we have LHS as $$\frac{3+\frac{\tan 10^{\circ}}{\tan 20^{\circ}}}{\tan 10^{\circ}+\frac{1}{\tan 20^{\circ}}}$$
assuming $y=10$ and using $\tan 2y=\frac{2 \tan y}{1-\tan ^2 y}$ we get
LHS as
$$\frac{3+\frac{\tan y}{\tan 2y}}{\tan y+\frac{1}{\tan 2y}}=\frac{\tan y(7-\tan ^2 y)}{1+\tan ^2 y}$$
how to proceed further?
 A: Let $\alpha = 20^\circ, \,\, u = \tan(\alpha)$, then
$$ LHS = \frac{1 + 3\tan(80^\circ)\tan(20^\circ)}{\tan(80^\circ) + \tan(20^\circ)} = \frac{1 + 3\tan(\alpha + 60^\circ)\tan(\alpha)}{\tan(\alpha + 60^\circ) + \tan(\alpha)} .$$
notice that 
$$ \tan(\alpha + 60^\circ) = \frac{\sqrt{3} + \tan{\alpha}}{1 - \sqrt{3} \tan{\alpha}} = \frac{\sqrt{3}+u}{1-u\sqrt{3}},$$
thus the LHS becomes
$$ LHS = \frac{1 + 3u\cdot \frac{\sqrt{3}+u}{1-u\sqrt{3}}}{u + \frac{\sqrt{3}+u}{1-u\sqrt{3}}} = \frac{1 - u\sqrt{3} + 3u\sqrt{3} + 3u^2}{u - u^2\sqrt{3}+ u + \sqrt{3}} = \frac{u\sqrt{3} + 1}{\sqrt{3} - u} = \cot(60^\circ - \alpha) = \cot(40^\circ).$$
A: Changing in terms of $$\sin \text {and} \cos ,3=2+1$$
$$L.H.S=\frac {2\sin 80^{\circ}.\sin20^{\circ} +\cos (80^{\circ}-20^{\circ})}{\sin(80^{\circ}+20^{\circ})} $$
$$=\frac {\cos(80^{\circ}-20^{\circ}) -\cos(80^{\circ}+20^{\circ})+\cos 60^{\circ}}{\sin 100^{\circ}} $$
$$=\frac {1-\cos 100^{\circ}}{\sin100^{\circ}}=\tan 50^{\circ}=\cot 40^{\circ}$$
A: If $\cot3A=\cot3x,3x=n180^\circ+3A$ where $n$ is any integer
$x=n60^\circ+A$ where $n\equiv0,1,2\pmod3$ 
$$\cot3A=\cot3x=\dfrac1{\tan3x}=\dfrac{1-3\tan^2x}{3\tan x-\tan^3x}=\dfrac{\cot^3x-3\cot x}{3\cot^2x-1}$$
$$\iff\cot^3x-3\cot3A\cot^2x-3\cot x+\cot3A=0\ \ \ \ (1)$$
If $\cot x_1,\cot x_2,\cot x_3$ are roots of $(1),$
$$\cot x_1\cot x_2+\cot x_2\cot x_3+\cot x_3\cot x_1=-\dfrac31$$
$$\implies\cot A\cot(A+60^\circ)+\cot(A+60^\circ)\cot(A+120^\circ)+\cot(A+120^\circ)\cot A=-3$$
Here $A=20^\circ$
