Prove that \((\sqrt{3}+1)(3-\cot 30^\circ)=\tan^3 60^\circ-2\sin 60^\circ\).
Both sides equal \(2\sqrt{3}\).
Step 1: Write down what is given.
We need to prove that
\[(\sqrt{3}+1)(3-\cot 30^\circ) = \tan^3 60^\circ - 2\sin 60^\circ.\]
Step 2: Recall trigonometric values.
Step 3: Simplify the Left-Hand Side (LHS).
LHS = \((\sqrt{3} + 1)(3 - \cot 30^\circ)\)
Substitute \(\cot 30^\circ = \sqrt{3}\):
LHS = \((\sqrt{3} + 1)(3 - \sqrt{3})\)
Now expand using distributive law:
= \( (\sqrt{3} \times 3) + (1 \times 3) - (\sqrt{3} \times \sqrt{3}) - (1 \times \sqrt{3}) \)
= \( 3\sqrt{3} + 3 - 3 - \sqrt{3} \)
= \( (3\sqrt{3} - \sqrt{3}) + (3 - 3) \)
= \( 2\sqrt{3} \)
So, LHS = \(2\sqrt{3}\).
Step 4: Simplify the Right-Hand Side (RHS).
RHS = \( \tan^3 60^\circ - 2\sin 60^\circ \)
Substitute values:
\( \tan 60^\circ = \sqrt{3} \Rightarrow \tan^3 60^\circ = (\sqrt{3})^3 = 3\sqrt{3} \)
\( \sin 60^\circ = \dfrac{\sqrt{3}}{2} \Rightarrow 2\sin 60^\circ = 2 \times \dfrac{\sqrt{3}}{2} = \sqrt{3} \)
So RHS = \( 3\sqrt{3} - \sqrt{3} = 2\sqrt{3} \)
Step 5: Compare both sides.
LHS = \(2\sqrt{3}\), RHS = \(2\sqrt{3}\)
Since both are equal, the given equation is proved.