Show that:

(i)

$\mathrm{LHS}=\frac{1-\sin 60^{\circ}}{\cos 60^{\circ}}=\frac{1-\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\frac{\left(\frac{2 \sqrt{3}}{2}\right)}{\frac{1}{2}}=\left(\frac{2-\sqrt{3}}{2}\right) \times 2=2-\sqrt{3}$

RHS $=\frac{\tan 60^{\circ}-1}{\tan 60^{\circ}+1}=\frac{\sqrt{3}-1}{\sqrt{3}+1}=\frac{\sqrt{3}-1}{\sqrt{3}+1} \times \frac{\sqrt{3}-1}{\sqrt{3}-1}=\frac{(\sqrt{3}-1)^{2}}{(\sqrt{3})^{2}-1^{2}}=\frac{3+1-2 \sqrt{3}}{3-1}=\frac{4-2 \sqrt{3}}{2}=2-\sqrt{3}$

Hence, LHS = RHS

$\therefore \frac{1-\sin 60^{\circ}}{\cos 60^{\circ}}=\frac{\tan 60^{\circ}-1}{\tan 60^{\circ}+1}$

(ii)

$\mathrm{LHS}=\frac{\cos 30^{\circ}+\sin 60^{\circ}}{1+\sin 30^{\circ}+\cos 60^{\circ}}=\frac{\left(\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right)}{\left(1+\frac{1}{2}+\frac{1}{2}\right)}=\frac{\frac{\sqrt{3}+\sqrt{3}}{2}}{\frac{2+1+1}{2}}=\frac{\sqrt{3}}{2}$

Also, RHS $=\cos 30^{\circ}=\frac{\sqrt{3}}{2}$

Hence, LHS = RHS

$\therefore \frac{\cos 30^{\circ}+\sin 60^{\circ}}{1+\sin 30^{\circ}+\cos 60^{\circ}}=\cos 30^{\circ}$