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

To do:

We have to prove that \( (\sqrt{3}+1)(3-\cot 30^{\circ})=\tan^{3} 60^{\circ}-2\sin 60^{\circ} \).

Solution:  

We know that,

$\cot 30^{\circ}=\sqrt3$

$\tan 60^{\circ}=\sqrt3$

$\sin 60^{\circ}=\frac{\sqrt3}{2}$

Let us consider LHS,

$(\sqrt{3}+1)(3-\cot 30^{\circ})=(\sqrt{3}+1)(3-\sqrt3)$

$=(\sqrt{3})(3)-(\sqrt3)^2+1(3)-1(\sqrt3)$

$=3\sqrt3-3+3-\sqrt3$

$=2\sqrt3$     

Let us consider RHS,

$\tan^{3} 60^{\circ}-2\sin 60^{\circ}=(\sqrt3)^3-2(\frac{\sqrt3}{2})$

$=3\sqrt3-\sqrt3$

$=2\sqrt3$

LHS $=$ RHS

Hence proved.

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Updated on: 10-Oct-2022

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