Question

Prove that $\cot \theta +\cot ({60}^o+\theta )+\cot ({120}^o+\theta )=3\cot 3\theta$

Answer 1

We know that,

$\cot (A+B)=\frac{\cot A\cdot \cot B-1}{\cot B+\cot A}$

So,

$L.H.S=\cot \theta +\cot ({60}^o+\theta )+\cot ({120}^o+\theta )=\cot \theta +\frac{\cot {60}^o\cdot \cot \theta -1}{\cot \theta +\cot {60}^o}+\frac{\cot {120}^o\cdot \cot \theta -1}{\cot \theta +\cot {120}^o}=\cot \theta +\frac{\frac{1}{\sqrt{3}}\cdot \cot \theta -1}{\cot \theta +\frac{1}{\sqrt{3}}}+\frac{-\frac{1}{\sqrt{3}}\cdot \cot \theta -1}{\cot \theta -\frac{1}{\sqrt{3}}}=\cot \theta +\frac{(\frac{1}{\sqrt{3}}\cot \theta -1)\cdot (\cot \theta -\frac{1}{\sqrt{3}})+(-\frac{1}{\sqrt{3}}\cot \theta -1)\cdot (\cot \theta +\frac{1}{\sqrt{3}})}{{\cot }^2\theta -\frac{1}{3}}=\cot \theta +\frac{\frac{1}{\sqrt{3}}{\cot }^2\theta -\cot \theta -\frac{1}{3}\cot \theta +\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{3}}{\cot }^2\theta -\cot \theta -\frac{1}{3}\cot \theta -\frac{1}{\sqrt{3}}}{{\cot }^2\theta -\frac{1}{3}}=\frac{{\cot }^3\theta -\frac{1}{3}\cot \theta +\frac{1}{\sqrt{3}}{\cot }^2\theta -\cot \theta -\frac{1}{3}\cot \theta +\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{3}}{\cot }^2\theta -\cot \theta -\frac{1}{3}\cot \theta -\frac{1}{\sqrt{3}}}{\frac{3{\cot }^2\theta -1}{3}}=\frac{{\cot }^3\theta -\frac{3}{3}\cot \theta -2\cot \theta }{\frac{3{\cot }^2\theta -1}{3}}=\frac{3({\cot }^3\theta -3\cot \theta )}{(3{\cot }^2\theta -1)}=3\cot 3\theta [\because \cot 3A=\frac{{\cot }^{3}A-3\cot A}{3{\cot }^{2}A-1}]=R.H.S$