Question

$\cot x+\cot \left(\frac{\pi }{3}+x\right)+\cot \left(\frac{\pi }{3}-x\right)=3\cot 3x$

Answer 1

$$\begin{gathered} \frac{\pi }{3}=60° \\ \mathrm{LHS}=\cot x+\cot \left(60°+x\right)-\cot \left(60°-x\right) \\ =\frac{1}{\tan x}+\frac{1}{\tan \left(60°+x\right)}-\frac{1}{\tan \left(60°-x\right)} \end{gathered}$$

$$\begin{gathered} =\frac{1}{\tan x}+\frac{1-\sqrt{3}\tan x}{\sqrt{3}+\tan x}-\frac{1+\sqrt{3}\tan x}{\sqrt{3}-\tan x} \\ \left[\tan \left(x+y\right)=\frac{\tan x+\tan y}{1-\tan x\tan y}\mathrm{and}\tan \left(x-y\right)=\frac{\tan x-\tan y}{1+\tan x\tan y}\right] \\ =\frac{1}{\tan x}-\frac{8\tan x}{3-{\tan }^{2}x} \\ =\frac{3-{\tan }^{2}x-8{\tan }^{2}x}{\left(\tan x\right)\left(3-{\tan }^{2}x\right)} \\ =\frac{3-9{\tan }^{2}x}{3\tan x-{\tan }^{3}x} \\ =3\left(\frac{1-3{\tan }^{2}x}{3\tan x-{\tan }^{3}x}\right) \\ =3\times \frac{1}{\tan 3x}\left(\because \tan 3\mathrm{\theta }=\frac{1-3{\tan }^2\mathrm{\theta }}{3\mathrm{tan\theta }-{\tan }^3\mathrm{\theta }}\right) \\ =3\cot 3x \\ =\mathrm{RHS} \\ \mathrm{Hence}\mathrm{proved}. \end{gathered}$$