Question

Question 5
$(\sqrt{3}+1)(3-cot{30}^{\circ })=ta{n}^3{60}^{\circ }-2sin{60}^{\circ }$

Answer 1

$RHS=ta{n}^3{60}^{\circ }-2sin{60}^{\circ }=(\sqrt{3}{)}^3-2\frac{\sqrt{3}}{2}=3\sqrt{3}-\sqrt{3}=2\sqrt{3}$

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

$[\because tan{60}^{\circ }=\sqrt{3},sin{60}^{\circ }=\frac{\sqrt{3}}{2}andcot{30}^{\circ }=\sqrt{3}]$

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

$\therefore LHS=RHS$