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Solution:

(i) $\mathrm{LHS}=2 \sin 50^{\circ} \cos 85^{\circ}$

$=\frac{\sin \left(50^{\circ}+85^{\circ}\right)+\sin \left(50^{\circ}-85^{\circ}\right)}{2}$   $\left[\because \sin A \cos B=\frac{1}{2}\{\sin (A+B)+\sin (A-B)\}\right]$

$=\frac{\sin 135^{\circ}+\sin \left(-35^{\circ}\right)}{2}$

$=\frac{\sin 135^{\circ}-\sin 35^{\circ}}{2}$

$=\frac{\cos 45^{\circ}-\sin 35^{\circ}}{2} \quad\left[\because \sin \left(90^{\circ}+45^{\circ}\right)=\cos 45^{\circ}\right]$

$=\frac{1}{2}\left(\frac{1}{\sqrt{2}}-\sin 35^{\circ}\right)$

$=\frac{1}{2}\left[\frac{1-\sqrt{2} \sin 35^{\circ}}{\sqrt{2}}\right]$

$=\frac{1-\sqrt{2} \sin 35^{\circ}}{2 \sqrt{2}}$

RHS $=\frac{1-\sqrt{2} \sin 35^{\circ}}{2 \sqrt{2}}$

Hence, LHS $=$ RHS

(ii) LHS $=2 \sin 25^{\circ} \cos 115^{\circ}$

$=\frac{\sin \left(25^{\circ}+115^{\circ}\right)+\sin \left(25^{\circ}-115^{\circ}\right)}{2} \quad\left[\because \sin A \cos B=\frac{1}{2}\{\sin (A+B)+\sin (A-B)\}\right]$

$=\frac{\sin 140^{\circ}+\sin \left(-90^{\circ}\right)}{2}$

$=\frac{\sin 140^{\circ}-\sin \left(90^{\circ}\right)}{2}$

$=\frac{\sin 140^{\circ}-1}{2}$

$\mathrm{RHS}=\frac{\sin 140^{\circ}-1}{2}$

Hence, LHS $=$ RHS