Question

$\cot A+\cot \left(60°+A\right)-cot\left(60°-A\right)=3\cot 3A$

Answer 1

$$\begin{gathered} \mathrm{LHS}=\mathrm{cotA}+\cot \left(60°+\mathrm{A}\right)-\cot \left(60°-\mathrm{A}\right) \\ =\frac{1}{\mathrm{tanA}}+\frac{1}{\tan \left(60°+\mathrm{A}\right)}-\frac{1}{\tan \left(60°-\mathrm{A}\right)} \end{gathered}$$

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