Prove that

(i) cos² 30° + sin 30° + tan² 45° = $2\dfrac{1}{4}$

(ii) 4(sin⁴ 30° + cos⁴ 60°) - 3(cos² 45° - sin² 90°) = 2

(iii) cos 60° = cos² 30° - sin² 30°.

(i) Solving,

L.H.S. of the equation : cos² 30° + sin 30° + tan² 45° = $2\dfrac{1}{4}$.

$\Rightarrow \Big(\dfrac{\sqrt{3}}{2}\Big)^2 + \dfrac{1}{2} + (1)^2 \\[1em] \Rightarrow \dfrac{3}{4} + \dfrac{1}{2} + 1 \\[1em] \Rightarrow \dfrac{3 + 2 + 4}{4} \\[1em] \Rightarrow \dfrac{9}{4} \\[1em] \Rightarrow 2\dfrac{1}{4}.$

Since, L.H.S. = R.H.S.

Hence, proved that cos² 30° + sin 30° + tan² 45° = $2\dfrac{1}{4}$.

(ii) Solving,

L.H.S. of the equation : 4(sin⁴ 30° + cos⁴ 60°) - 3(cos² 45° - sin² 90°) = 2

$\Rightarrow 4\Big[\Big(\dfrac{1}{2}\Big)^4 + \Big(\dfrac{1}{2}\Big)^4\Big] - 3\Big[\Big(\dfrac{1}{\sqrt{2}}\Big)^2 - (1)^2\Big] \\[1em] \Rightarrow 4\Big[\dfrac{1}{16} + \dfrac{1}{16}\Big] - 3\Big[\dfrac{1}{2} - 1\Big] \\[1em] \Rightarrow 4 \times \dfrac{2}{16} - 3 \times -\dfrac{1}{2} \\[1em] \Rightarrow \dfrac{1}{2} + \dfrac{3}{2} \\[1em] \Rightarrow \dfrac{4}{2} \\[1em] \Rightarrow 2.$

Since, L.H.S. = R.H.S.

Hence, proved that 4(sin⁴ 30° + cos⁴ 60°) - 3(cos² 45° - sin² 90°) = 2.

(iii) Solving,

R.H.S. of the equation : cos 60° = cos² 30° - sin² 30°.

$\Rightarrow \Big(\dfrac{\sqrt{3}}{2}\Big)^2 - \Big(\dfrac{1}{2}\Big)^2 \\[1em] \Rightarrow \dfrac{3}{4} - \dfrac{1}{4} \\[1em] \Rightarrow \dfrac{2}{4} \\[1em] \Rightarrow \dfrac{1}{2} \\[1em] \Rightarrow \text{cos 60°}.$

Since, L.H.S. = R.H.S.

Hence, proved that cos 60° = cos² 30° - sin² 30°.