Question

Prove that $\left(\sqrt{3}+1\right)\left(3-cot{30}^{°}\right)={\tan }^3{60}^{°}-2\sin {60}^{°}$ .

Answer 1

Consider the left hand side.
$$\begin{gathered} \left(\sqrt{3}+1\right)\left(3-\cot 30°\right) \\ =\left(\sqrt{3}+1\right)\left(3-\sqrt{3}\right)\left(\cot 30°=\sqrt{3}\right) \\ =\left(\sqrt{3}+1\right)\times \sqrt{3}\times \left(\sqrt{3}-1\right) \\ =\left[{\left(\sqrt{3}\right)}^2-1\right]\times \sqrt{3} \\ =2\sqrt{3} \end{gathered}$$
Now, consider the right hand side.
$$\begin{gathered} {\tan }^{3}60°-2\sin 60° \\ ={\left(\sqrt{3}\right)}^3-\left(2\times \frac{\sqrt{3}}{2}\right)\left(\sin 60°=\frac{\sqrt{3}}{2}\right) \\ =\sqrt{3}\times \left[{\left(\sqrt{3}\right)}^2-1\right] \\ =2\sqrt{3} \end{gathered}$$
So, LHS = RHS
Hence proved.