Find the value of :

3 sin² 30° + 2 tan² 60° - 5 cos² 45°

3 sin² 30° + 2 tan² 60° - 5 cos² 45°

$= 3 \times \Big(\dfrac{1}{2}\Big)^2 + 2 \times (\sqrt3)^2 - 5 \times \Big(\dfrac{1}{\sqrt2}\Big)^2\\[1em] = 3 \times \dfrac{1}{4} + 2 \times 3 - 5 \times \dfrac{1}{2}\\[1em] = \dfrac{3}{4} + 6 - \dfrac{5}{2}\\[1em] = \dfrac{3}{4} + \dfrac{6 \times 4}{4} - \dfrac{5 \times 2}{2 \times 2}\\[1em] = \dfrac{3}{4} + \dfrac{24}{4} - \dfrac{10}{4}\\[1em] = \dfrac{3 + 24 - 10}{4}\\[1em] = \dfrac{17}{4}\\[1em] = 4\dfrac{1}{4}$

Hence, 3 sin² 30° + 2 tan² 60° - 5 cos² 45° = $4\dfrac{1}{4}$.