Find the values of

(i) $\dfrac{\text{sin 60°}}{\text{cos}^2 45°} - 3\text{ tan 30° + 5 cos 90°}$

(ii) $2\sqrt{2}\text{ cos 45° cos 60°} + 2\sqrt{3} \text{ sin 30° tan 60° - cos 0°}$

(iii) $\dfrac{4}{5}\text{tan}^2 60° - \dfrac{2}{\text{sin}^2 30°} - \dfrac{3}{4}\text{tan}^2 30°$

(i) Solving,

$\Rightarrow \dfrac{\dfrac{\sqrt{3}}{2}}{\Big(\dfrac{1}{\sqrt{2}}\Big)^2} - 3 \times \dfrac{1}{\sqrt{3}} + 5 \times 0 \\[1em] \Rightarrow \dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} - \sqrt{3} \\[1em] \Rightarrow \dfrac{2\sqrt{3}}{2} - \sqrt{3} \\[1em] \Rightarrow \sqrt{3} - \sqrt{3} \\[1em] \Rightarrow 0.$

Hence, $\dfrac{\text{sin 60°}}{\text{cos}^2 45°} - 3\text{ tan 30° + 5 cos 90°} = 0.$

(ii) Solving,

$\Rightarrow 2\sqrt{2}\text{ cos 45° cos 60°} + 2\sqrt{3} \text{ sin 30° tan 60° - cos 0°} \\[1em] \Rightarrow 2\sqrt{2} \times \dfrac{1}{\sqrt{2}} \times \dfrac{1}{2} + 2 \times \sqrt{3} \times \dfrac{1}{2} \times \sqrt{3} - 1 \\[1em] \Rightarrow 1 + \sqrt{3} \times \sqrt{3} - 1 \\[1em] \Rightarrow 1 + 3 - 1 \\[1em] \Rightarrow 3.$

Hence, $2\sqrt{2}\text{ cos 45° cos 60°} + 2\sqrt{3} \text{ sin 30° tan 60° - cos 0°} = 3.$

(iii) Solving,

$\Rightarrow \dfrac{4}{5}\text{tan}^2 60° - \dfrac{2}{\text{sin}^2 30°} - \dfrac{3}{4}\text{tan}^2 30° \\[1em] \Rightarrow \dfrac{4}{5} \times (\sqrt{3})^2 - \dfrac{2}{\Big(\dfrac{1}{2}\Big)^2} - \dfrac{3}{4} \times \Big(\dfrac{1}{\sqrt{3}}\Big)^2 \\[1em] \Rightarrow \dfrac{4}{5} \times 3 - 2 \times (2)^2 - \dfrac{3}{4} \times \dfrac{1}{3} \\[1em] \Rightarrow \dfrac{12}{5} - 8 - \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{48 - 160 - 5}{20} \\[1em] \Rightarrow -\dfrac{117}{20}\\[1em] \Rightarrow -5\dfrac{17}{20}.$

Hence, $\dfrac{4}{5}\text{tan}^2 60° - \dfrac{2}{\text{sin}^2 30°} - \dfrac{3}{4}\text{tan}^2 30° = -5\dfrac{17}{20}$.