Without using trigonometric tables, find the values of;

(i) $\dfrac{\sin 30^\circ - \sin 90^\circ +2\times \cos 0^\circ}{\tan 30^\circ \times \tan 60^\circ}$

(ii) $\dfrac{5 \sin^2 30^\circ + \cos^2 45^\circ- 4 \tan^2 30^\circ}{2 \sin 30^\circ \cos 30^\circ+ \tan 45^\circ}$

(iii) $\dfrac{\tan^2 60^\circ + 4 \cos^2 45^\circ + \sec^2 30^\circ + 5 \cos^2 90^\circ}{\cosec 30^\circ + \sec 60^\circ - \cot^2 30^\circ}$

(iv) 4(sin⁴ 30° + cos⁴ 60°) - 3 (cos² 45° - sin² 90°)

(i) Solving,

$\dfrac{\sin 30^\circ - \sin 90^\circ +2\times \cos 0^\circ}{\tan 30^\circ \times \tan 60^\circ}\\[1em] = \dfrac{\dfrac{1}{2} - 1 +2\times1}{\dfrac{1}{\sqrt{3}}\times {\sqrt{3}}}\\[1em] = \dfrac{\dfrac{1}{2} - 1 + 2}{\dfrac{1}{\sqrt{3}}\times {\sqrt{3}}}\\[1em] = \dfrac{\dfrac{1}{2} + 1}{1}\\[1em] = \dfrac{3}{2}.$

Hence, the required value = $\dfrac{3}{2}$.

(ii) As,

sin² 30° = (sin 30°)² = $\Big(\dfrac{1}{2}\Big)^2 = \dfrac{1}{4}$

cos² 45° = (cos 45°)² = $\Big(\dfrac{1}{\sqrt{2}}\Big)^2 = \dfrac{1}{2}$

tan² 30° = (tan 30°)² = $\Big(\dfrac{1}{\sqrt{3}}\Big)^2 = \dfrac{1}{3}$

Therefore,

$\dfrac{5 \sin^2 30^\circ + \cos^2 45^\circ- 4 \tan^2 30^\circ}{2 \sin 30^\circ \cos 30^\circ+ \tan 45^\circ}\\[1em] =\dfrac{5\times\dfrac{1}{4} + \dfrac{1}{2} - 4\times\dfrac{1}{3}}{2\times\dfrac{1}{2} \times\dfrac{\sqrt{3}}{2}+ 1 }\\[1em] = \dfrac{\dfrac{5}{4} + \dfrac{1}{2} -\dfrac{4}{3}}{\dfrac{\sqrt{3}}{2}+1}\\[1em] = \dfrac{\dfrac{15 + 6 - 16}{12}}{\dfrac{\sqrt{3}}{2}+1}\\[1em] = \dfrac{\dfrac{5}{12}}{\dfrac{\sqrt{3} + 2}{2}} \\[1em] = \dfrac{5}{6(2 + \sqrt{3})}.$

Hence, $\dfrac{5 \sin^2 30^\circ + \cos^2 45^\circ- 4 \tan^2 30^\circ}{2 \sin 30^\circ \cos 30^\circ+ \tan 45^\circ} = \dfrac{5}{6(2 + \sqrt{3})}$.

(iii) As,

tan² 60° = (tan 60°)² = ($\sqrt{3}$)² = 3

cos² 45° = (cos 45°)² = $\Big(\dfrac{1}{\sqrt{2}}\Big)^2 = \dfrac{1}{2}$

sec² 30° = (sec 30°)² = $\Big(\dfrac{2}{\sqrt{3}}\Big)^2 = \dfrac{4}{3}$

cos² 90° = (cos 90°)² = 0

cot² 30° = (cot 30°)² = ($\sqrt{3}$)² = 3

Therefore,

$\dfrac{\tan^2 60^\circ + 4 \cos^2 45^\circ + \sec^2 30^\circ + 5 \cos^2 90^\circ}{\cosec 30^\circ + \sec 60^\circ - \cot^2 30^\circ}\\[1em] = \dfrac{3 + 4\times\dfrac{1}{2} + \dfrac{4}{3} + 0}{2 + 2 - 3}\\[1em] = \dfrac{3 + 2 + \dfrac{4}{3} + 0}{1}\\[1em] = 5+ \dfrac{4}{3}\\[1em] = \dfrac{19}{3} \\[1em] = 6\dfrac{1}{3}.$

Hence, $\dfrac{\tan^2 60^\circ + 4 \cos^2 45^\circ + \sec^2 30^\circ + 5 \cos^2 90^\circ}{\cosec 30^\circ + \sec 60^\circ - \cot^2 30^\circ} = \dfrac{19}{3}$

(iv) As,

sin⁴ 30° = (sin 30°)⁴ = $\Big(\dfrac{1}{2}\Big)^4 = \dfrac{1}{16}$

cos⁴ 60° = (cos 60°)⁴ = $\Big(\dfrac{1}{2}\Big)^4 = \dfrac{1}{16}$

cos² 45° = (cos 45°)² = $\Big(\dfrac{1}{\sqrt{2}}\Big)^2 = \dfrac{1}{2}$

sin² 90° = (sin 90°)² = 1

Therefore,

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

= $4\Big(\dfrac{1}{16} + \dfrac{1}{16}\Big) - 3 \Big(\dfrac{1}{2} - 1\Big)$

= $4\Big(\dfrac{2}{16}\Big) - 3 \Big(\dfrac{-1}{2}\Big)$

= $\dfrac{1}{2} + \dfrac{3}{2}$

= 2.

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