Question

Prove that following identities:

$cotA+cot({60}^{\circ }+A)+cot({120}^{\circ }+A)=3cot3A$

Answer 1

$LHS=cotA+cot({60}^{\circ }+A)+cot({120}^{\circ }+A)$

$cotA+cot({60}^{\circ }+A)-cot[{180}^{\circ }-({120}^{\circ }+A\left)\right]\{\text{Since}-cot\theta -cot({180}^{\circ }-\theta \left)\right\}=cotA+cot({60}^{\circ }+A)-cot({60}^{\circ }-A)=\frac{1}{tanA}+\frac{1}{tan({60}^{\circ }+A)}-\frac{1}{tan({60}^{\circ }-A)}=\frac{1}{tanA}+\frac{1-\sqrt{3}tanA}{\sqrt{3}+tanA}-\frac{1+\sqrt{3}tanA}{\sqrt{3}-tanA}=\frac{1}{tanA}-\frac{8tanA}{3-ta{n}^{2}A}=\frac{3-ta{n}^{2}A-8ta{n}^{2}A}{3tanA-ta{n}^{3}A}=\frac{3-9ta{n}^{2}A}{3tanA-ta{n}^{3}A}=\frac{3(1-3ta{n}^{2}A)}{3tanA-ta{n}^{3}A}=\frac{3}{tan3A}=3cot3A$