Compound angle formula Vs Triple angle

 Prove that $ \displaystyle \cot (\alpha +\beta )=\frac{{\cot \alpha \cot \beta -1}}{{\cot \alpha +\cot \beta }}$ and hence prove that $ \displaystyle \cot \theta +\cot ({{60}^{\circ }}+\theta )-\cot ({{60}^{\circ }}-\theta )=3\cot 3\theta $ .


$ \displaystyle \cot (\alpha +\beta )=\frac{1}{{\tan (\alpha +\beta )}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{1-\tan \alpha \tan \beta }}{{\tan \alpha +\tan \beta }}$

$ \displaystyle \ \ \ \ \ \ \ \,\ \ \ \ \ \ \ \ =\frac{{1-\frac{1}{{\cot \alpha \cot \beta }}}}{{\frac{1}{{\cot \alpha }}+\frac{1}{{\cot \beta }}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\frac{{\cot \alpha \cot \beta -1}}{{\cot \alpha \cot \beta }}}}{{\frac{{\cot \beta +\cot \alpha }}{{\cot \alpha \cot \beta }}}}$

$ \displaystyle \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\frac{{\cot \alpha \cot \beta -1}}{{\cot \beta +\cot \alpha }}$

$ \displaystyle \cot \theta +\cot ({{60}^{\circ }}+\theta )-\cot ({{60}^{\circ }}-\theta )$

$ \displaystyle =\cot \theta +\frac{{\cot {{{60}}^{\circ }}\cot \theta -1}}{{\cot {{{60}}^{\circ }}+\cot \theta }}-\frac{{\cot {{{60}}^{\circ }}\cot \theta +1}}{{\cot \theta -\cot {{{60}}^{\circ }}}}$

$ \displaystyle =\cot \theta +\frac{{\frac{1}{{\sqrt{3}}}\cot \theta -1}}{{\frac{1}{{\sqrt{3}}}+\cot \theta }}-\frac{{\frac{1}{{\sqrt{3}}}\cot \theta +1}}{{\cot \theta -\frac{1}{{\sqrt{3}}}}}$

$ \displaystyle =\cot \theta +\frac{{\cot \theta -\sqrt{3}}}{{1+\sqrt{3}\cot \theta }}-\frac{{\cot \theta +\sqrt{3}}}{{\sqrt{3}\cot \theta -1}}$

$ \displaystyle =\cot \theta +\frac{{\sqrt{3}{{{\cot }}^{2}}\theta -\cot \theta -3\cot \theta +\sqrt{3}-\cot \theta -\sqrt{3}-\sqrt{3}{{{\cot }}^{2}}\theta -3\cot \theta }}{{3{{{\cot }}^{2}}\theta -1}}$

$ \displaystyle =\cot \theta -\frac{{8\cot \theta }}{{3{{{\cot }}^{2}}\theta -1}}$

$ \displaystyle =\frac{{3{{{\cot }}^{3}}\theta -9\cot \theta }}{{3{{{\cot }}^{2}}\theta -1}}$

$ \displaystyle \cot 2\theta =\frac{{{{{\cot }}^{2}}\theta -1}}{{2\cot \theta }}$

$ \displaystyle \cot 3\theta =\cot (2\theta +\theta )$

$ \displaystyle \ \ \ \ \ \ \ \ \ =\frac{{\cot 2\theta \cot \theta -1}}{{\cot 2\theta +\cot \theta }}$

$ \displaystyle \ \ \ \ \ \ \ \ \ =\frac{{(\frac{{{{{\cot }}^{2}}\theta -1}}{{2\cot \theta }})\cot \theta -1}}{{\frac{{{{{\cot }}^{2}}\theta -1}}{{2\cot \theta }}+\cot \theta }}$

$ \displaystyle \ \ \ \ \ \ \ \ \ =\frac{{{{{\cot }}^{3}}\theta -\cot \theta -2\cot \theta }}{{{{{\cot }}^{2}}\theta -1+2{{{\cot }}^{2}}\theta }}$

$ \displaystyle \ \ \ \ \ \ \ \ \ =\frac{{{{{\cot }}^{3}}\theta -3\cot \theta }}{{3{{{\cot }}^{2}}\theta -1}}$

$ \displaystyle 3\cot 3\theta =\frac{{3{{{\cot }}^{3}}\theta -9\cot \theta }}{{3{{{\cot }}^{2}}\theta -1}}$

$ \displaystyle \therefore \cot \theta +\cot ({{60}^{\circ }}+\theta )-\cot ({{60}^{\circ }}-\theta )=3\cot 3\theta $