Question

Prove the following: $(\sqrt{3}+1)(3-\cot 30°)={\tan }^{3}60°-2\sin 60°$

Answer 1

Step 1. Evaluate LHS of the given expression.

$$\begin{gathered} \mathrm{LHS}=(\sqrt{3}+1)(3-\cot 30°) \\ =(\sqrt{3}+1)\left(3-\sqrt{3}\right)\left[\because \cot 30°=\sqrt{3}\right] \\ =3\sqrt{3}-3+3-\sqrt{3} \\ =2\sqrt{3} \end{gathered}$$

Step 2. Evaluate RHS of the given expression.

$$\begin{gathered} \mathrm{RHS}={\tan }^{3}60°-2\sin 60° \\ ={\left(\sqrt{3}\right)}^3-2·\frac{\sqrt{3}}{2}\left[\because \tan 60°=\sqrt{3},\sin 60°=\frac{\sqrt{3}}{2}\right] \\ =3\sqrt{3}-\sqrt{3} \\ =2\sqrt{3} \end{gathered}$$

$\mathrm{LHS}=\mathrm{RHS}$

Hence, proved.