Prove that:

(i) $4 \pi=720^{\circ}, \frac{3 \pi}{2}=270^{\circ}, \frac{5 \pi}{6}=150^{\circ}, \frac{2 \pi}{3}=120^{\circ}$

$\mathrm{LHS}=\tan \left(720^{\circ}\right)-\cos \left(270^{\circ}\right)-\sin \left(150^{\circ}\right) \cos \left(120^{\circ}\right)$

$=\tan \left(90^{\circ} \times 8+0^{\circ}\right)-\cos \left(90^{\circ} \times 3+0^{\circ}\right)-\sin \left(90^{\circ} \times 1+60^{\circ}\right) \cos \left(90^{\circ} \times 1+30^{\circ}\right)$

$=\tan \left(0^{\circ}\right)-\sin \left(0^{\circ}\right)-\cos \left(60^{\circ}\right)\left[-\sin \left(30^{\circ}\right)\right]$

$=\tan \left(0^{\circ}\right)-\sin \left(0^{\circ}\right)+\cos \left(60^{\circ}\right) \sin \left(30^{\circ}\right)$

$=0-0+\frac{1}{2} \times \frac{1}{2}$

$=\frac{1}{4}$

= RHS 

Hence proved.

(ii) $\frac{13 \pi}{3}=780^{\circ}, \frac{8 \pi}{3}=480^{\circ}, \frac{2 \pi}{3}=120^{\circ}, \frac{5 \pi}{6}=150^{\circ}$

$=\sin \left(90^{\circ} \times 8+60^{\circ}\right) \sin \left(90^{\circ} \times 5+30^{\circ}\right)+\cos \left(90^{\circ} \times 1+30^{\circ}\right) \sin \left(90^{\circ} \times 1+60^{\circ}\right)$

$=\sin \left(60^{\circ}\right) \cos \left(30^{\circ}\right)+\left[-\sin \left(30^{\circ}\right)\right] \cos \left(60^{\circ}\right)$

$=\sin \left(60^{\circ}\right) \cos \left(30^{\circ}\right)-\sin \left(30^{\circ}\right) \cos \left(60^{\circ}\right)$

$=\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}-\frac{1}{2} \times \frac{1}{2}$

$=\frac{3}{4}-\frac{1}{4}$

$=\frac{1}{2}$

= RHS 

Hence proved.

(iii) $\frac{13 \pi}{2}=780^{\circ}, \frac{2 \pi}{a}=120^{\circ}, \frac{4 \pi}{2}=240^{\circ}, \frac{13 \pi}{c}=390^{\circ}$

LHS $=\sin \left(780^{\circ}\right) \sin \left(120^{\circ}\right)+\cos \left(240^{\circ}\right) \sin \left(390^{\circ}\right)$

$=\sin \left(90^{\circ} \times 8+60^{\circ}\right) \sin \left(90^{\circ} \times 1+30^{\circ}\right)+\cos \left(90^{\circ} \times 2+60^{\circ}\right) \sin \left(90^{\circ} \times 4+30^{\circ}\right)$

$=\sin 60^{\circ} \cos 30^{\circ}+\left[-\cos 60^{\circ}\right] \sin 30^{\circ}$

$=\sin 60^{\circ} \cos 30^{\circ}-\cos 60^{\circ} \sin 30^{\circ}$

$=\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}-\frac{1}{2} \times \frac{1}{2}$

$=\frac{3}{4}-\frac{1}{4}$

$=\frac{1}{2}$

= RHS 

Hence proved.

(iv) $\frac{10 \pi}{3}=600^{\circ}, \frac{13 \pi}{6}=390^{\circ}, \frac{8 \pi}{3}=480^{\circ}, \frac{5 \pi}{6}=150^{\circ}$

$\mathrm{LHS}=\sin 600^{\circ} \cos 390^{\circ}+\cos 480^{\circ} \sin 150^{\circ}$

$=\sin \left(90^{\circ} \times 6+60^{\circ}\right) \cos \left(90^{\circ} \times 4+30^{\circ}\right)+\cos \left(90^{\circ} \times 5+30^{\circ}\right) \sin \left(90^{\circ} \times 1+60^{\circ}\right)$

$=\left[-\sin 60^{\circ}\right] \cos 30^{\circ}+\left[-\sin 30^{\circ}\right] \cos 60^{\circ}$

$=-\sin 60^{\circ} \cos \left(30^{\circ}\right)-\sin 30^{\circ} \cos 60^{\circ}$

$=-\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}-\frac{1}{2} \times \frac{1}{2}$

$=-\frac{3}{4}-\frac{1}{4}$

$=-1$

= RHS 

Hence proved.

(v) $\frac{5 \pi}{4}=225^{\circ}, \frac{9 \pi}{4}=405^{\circ}, \frac{17 \pi}{4}=765^{\circ}, \frac{15 \pi}{4}=675^{\circ}$

$\mathrm{LHS}=\tan 225^{\circ} \cot 405^{\circ}+\tan 765^{\circ} \cot 675^{\circ}$

$=\tan \left(90^{\circ} \times 2+45^{\circ}\right) \cot \left(90^{\circ} \times 4+45^{\circ}\right)+\tan \left(90^{\circ} \times 8+45^{\circ}\right) \cot \left(90^{\circ} \times 7+45^{\circ}\right)$

$=\tan 45^{\circ} \cot 45^{\circ}+\tan 45^{\circ}\left[-\tan 45^{\circ}\right]$

$=\tan 45^{\circ} \cot 45^{\circ}-\tan 45^{\circ} \tan 45^{\circ}$

$=1 \times 1-1 \times 1$

$=1-1$

$=0$

= RHS 

Hence proved.