Question

$Prove that-- (\sqrt{3} +1)(3-tan60) = cot^{3} 30-2cos30$

Answer 1

$\large\underline{\bold{Given \:Question - }}$

$\sf \: Prove \: that \: ( \sqrt{3} + 1)(3 - tan60 \degree) = {cot}^{3}30\degree - 2cos30\degree$

$\large\underline{\sf{Solution-}}$

We know that

$\boxed{ \rm \: tan60\degree = \sqrt{3} }$

$\boxed{ \rm \: cot30\degree = \sqrt{3} }$

$\boxed{ \rm \: cos30\degree =\dfrac{ \sqrt{3} }{2}}$

Now,

Consider LHS,

$\rm :\longmapsto\:( \sqrt{3} + 1)(3 - tan60\degree)$

$\: \: = \sf \: ( \sqrt{3} + 1)(3 - \sqrt{3} )$

$\: \: = \sf \: ( \sqrt{3} + 1)( \sqrt{3} \times \sqrt{3} - \sqrt{3})$

$\: \: = \sf \: ( \sqrt{3} + 1) \times \sqrt{3} \times ( \sqrt{3} - 1)$

$\: \: = \sf \: \sqrt{3} \bigg( {( \sqrt{3}) }^{2} - {(1)}^{2} \bigg)$

$\: \: = \sf \: \sqrt{3} \times (3 - 1)$

$\: \: = \sf \: 2 \sqrt{3} - - - (1)$

Now,

Consider RHS,

$\rm :\longmapsto\: {cot}^{3}30\degree - 2cos30\degree$

$\: \: = \sf \: {\bigg( \sqrt{3} \bigg) }^{3} - 2 \times \dfrac{ \sqrt{3} }{2}$

$\: \: = \sf \: 3 \sqrt{3} - \sqrt{3}$

$\: \: = \sf \: 2 \sqrt{3} - - - (2)$

From equation (1) and equation (2), we concluded,

$\bf \: ( \sqrt{3} + 1)(3 - tan60 \degree) = {cot}^{3}30\degree - 2cos30\degree$

${\boxed{\boxed{\bf{Hence, Proved}}}}$

─━─━─━─━─━─━─━─━─━─━─━─━─

$\begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\sf Trigonometry\: Table \\ \begin{gathered}\begin{gathered}\begin{gathered}\begin{gathered}\boxed{\boxed{\begin{array}{ |c |c|c|c|c|c|} \bf\angle A & \bf{0}^{ \circ} & \bf{30}^{ \circ} & \bf{45}^{ \circ} & \bf{60}^{ \circ} & \bf{90}^{ \circ} \\ \\ \rm sin A & 0 & \dfrac{1}{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} &1 \\ \\ \rm cos \: A & 1 & \dfrac{ \sqrt{3} }{2}& \dfrac{1}{ \sqrt{2} } & \dfrac{1}{2} &0 \\ \\ \rm tan A & 0 & \dfrac{1}{ \sqrt{3} }&1 & \sqrt{3} & \rm \infty \\ \\ \rm cosec A & \rm \infty & 2& \sqrt{2} & \dfrac{2}{ \sqrt{3} } &1 \\ \\ \rm sec A & 1 & \dfrac{2}{ \sqrt{3} }& \sqrt{2} & 2 & \rm \infty \\ \\ \rm cot A & \rm \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } & 0\end{array}}}\end{gathered}\end{gathered}\end{gathered} \end{gathered}\end{gathered}\end{gathered}\end{gathered}\end{gathered}$