Prove that:

(i) LHS $=\tan 225^{\circ} \cot 405^{\circ}+\tan 765^{\circ} \cot 675^{\circ}$

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

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

$=1 \times 1+1 \times(-1)$

$=1-1$

$=0$

= RHS

Hence proved.

(ii) LHS $=\sin \frac{8 \pi}{3} \cos \frac{23 \pi}{6}+\cos \frac{13 \pi}{3} \sin \frac{35 \pi}{6}$

$=\sin \left(\frac{8}{3} \times 180^{\circ}\right) \cos \left(\frac{23}{6} \times 180^{\circ}\right)+\cos \left(\frac{13}{3} \times 180^{\circ}\right) \sin \left(\frac{35}{6} \times 180^{\circ}\right)$

$=\sin \left(480^{\circ}\right) \cos \left(690^{\circ}\right)+\cos \left(780^{\circ}\right) \sin \left(1050^{\circ}\right)$

$=\sin \left(90^{\circ} \times 5+30^{\circ}\right) \cos \left(90^{\circ} \times 7+60^{\circ}\right)+\cos \left(90^{\circ} \times 8+60^{\circ}\right) \sin \left(90^{\circ} \times 11+60^{\circ}\right)$

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

$=\frac{\sqrt{3}}{2} \times \frac{\sqrt{3}}{2}+\frac{1}{2} \times\left(-\frac{1}{2}\right)$

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

$=\frac{2}{4}$

$=\frac{1}{2}$

= RHS

Hence proved.

(iii) $\mathrm{LHS}=\cos 24^{\circ}+\cos 55^{\circ}+\cos 125^{\circ}+\cos 204^{\circ}+\cos 300^{\circ}$

$=\cos 24^{\circ}+\cos \left(90^{\circ}-35^{\circ}\right)+\cos \left(90^{\circ} \times 1+35^{\circ}\right)+\cos \left(90^{\circ} \times 2+24^{\circ}\right)+\cos$$\left(90^{\circ} \times 3+30^{\circ}\right)$

$=\cos 24^{\circ}+\sin 35^{\circ}-\sin 35^{\circ}-\cos 24^{\circ}+\sin 30^{\circ}$

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

$=\frac{1}{2}$

= RHS

Hence proved.

(iv) LHS $=\tan \left(-225^{\circ}\right) \cot \left(-405^{\circ}\right)-\tan \left(-765^{\circ}\right) \cot \left(675^{\circ}\right)$

$=\left[-\tan \left(225^{\circ}\right)\right]\left[-\cot \left(405^{\circ}\right)\right]-\left[-\tan \left(765^{\circ}\right)\right] \cot \left(675^{\circ}\right)$

$[\because \tan (-x)=\tan (x)$ and $\cot (-x)=-\cot (x)]$

$=\tan \left(225^{\circ}\right) \cot \left(405^{\circ}\right)+\tan \left(765^{\circ}\right) \cot \left(675^{\circ}\right)$

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

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

$=1 \times 1+1 \times(-1)$

$=1-1$

$=0$

= RHS

Hence proved.

(v) $\mathrm{LHS}=\cos \left(570^{\circ}\right) \sin \left(510^{\circ}\right)+\sin \left(-330^{\circ}\right) \cos \left(-390^{\circ}\right)$

$=\cos \left(570^{\circ}\right) \sin \left(510^{\circ}\right)+\left[-\sin \left(330^{\circ}\right)\right] \cos \left(390^{\circ}\right)$

$[\because \sin (-x)=-\sin x$ and $\cos (-x)=\cos x]$

$=\cos \left(570^{\circ}\right) \sin \left(510^{\circ}\right)-\sin \left(330^{\circ}\right) \cos \left(390^{\circ}\right)$

$=\cos \left(90^{\circ} \times 6+30^{\circ}\right) \sin \left(90^{\circ} \times 5+60^{\circ}\right)-\sin \left(90^{\circ} \times 3+60^{\circ}\right) \cos \left(90^{\circ} \times 4+30^{\circ}\right)$

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

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

$=0$

= RHS

Hence proved.

(vi) LHS $=\tan \frac{11 \pi}{3}-2 \sin \frac{4 \pi}{6}-\frac{3}{4} \operatorname{cosec}^{2} \frac{\pi}{4}+4 \cos ^{2} \frac{17 \pi}{6}$

$=\tan \left(\frac{11 \pi}{3}\right)-2 \sin \left(\frac{4 \pi}{6}\right)-\frac{3}{4}\left[\operatorname{cosec}\left(\frac{\pi}{4}\right)\right]^{2}+4\left[\cos \left(\frac{17 \pi}{6}\right)\right]^{2}$

$=\tan \left(\frac{11}{3} \times 180^{\circ}\right)-2 \sin \left(\frac{4}{6} \times 180^{\circ}\right)-\frac{3}{4}\left[\operatorname{cosec}\left(\frac{180^{\circ}}{4}\right)\right]^{2}+4\left[\cos \left(\frac{17 \times 180^{\circ}}{6}\right)\right]^{2}$

$=\tan \left(660^{\circ}\right)-2 \sin \left(120^{\circ}\right)-\frac{3}{4}\left[\operatorname{cosec}\left(45^{\circ}\right)\right]^{2}+4\left[\cos \left(510^{\circ}\right)\right]^{2}$

$=\tan \left(660^{\circ}\right)-2 \sin \left(120^{\circ}\right)-\frac{3}{4}\left[\operatorname{cosec}\left(45^{\circ}\right)\right]^{2}+4\left[\cos \left(510^{\circ}\right)\right]^{2}$

$=\tan \left(90^{\circ} \times 7+30^{\circ}\right)-2 \sin \left(90^{\circ} \times 1+30^{\circ}\right)-\frac{3}{4}\left[\operatorname{cosec}\left(45^{\circ}\right)\right]^{2}+4\left[\cos \left(90^{\circ} \times 5+60^{\circ}\right)\right]^{2}$

$=\left[-\cot \left(30^{\circ}\right)\right]-\left[2 \cos \left(30^{\circ}\right)\right]-\frac{3}{4}\left[\operatorname{cosec}\left(45^{\circ}\right)\right]^{2}+4\left[-\sin \left(60^{\circ}\right)\right]^{2}$

$=-\cot \left(30^{\circ}\right)-2 \cos \left(30^{\circ}\right)-\frac{3}{4}\left[\operatorname{cosec}\left(45^{\circ}\right)\right]^{2}+4\left[\sin \left(60^{\circ}\right)\right]^{2}$

$=-\sqrt{3}-\frac{2 \sqrt{3}}{2}-\frac{3}{4}[\sqrt{2}]^{2}+4\left[\frac{\sqrt{3}}{2}\right]^{2}$

$=-\sqrt{3}-\sqrt{3}-\frac{3}{2}+3$

$=\frac{3-4 \sqrt{3}}{2}$

= RHS

Hence proved.

(vii) LHS $=3 \sin \frac{\pi}{6} \sec \frac{\pi}{3}-4 \sin \frac{5 \pi}{6} \cot \frac{\pi}{4}$

$=3 \sin \left(\frac{180^{\circ}}{6}\right) \sec \left(\frac{180^{\circ}}{3}\right)-4 \sin \left(\frac{5 \times 180^{\circ}}{6}\right) \cot \left(\frac{180^{\circ}}{4}\right)$

$=3 \sin \left(30^{\circ}\right) \sec \left(60^{\circ}\right)-4 \sin \left(150^{\circ}\right) \cot \left(45^{\circ}\right)$

$=3 \sin \left(30^{\circ}\right) \sec \left(60^{\circ}\right)-4 \sin \left(90^{\circ} \times 1+60^{\circ}\right) \cot \left(45^{\circ}\right)$

$=3 \sin \left(30^{\circ}\right) \sec \left(60^{\circ}\right)-4 \cos \left(60^{\circ}\right) \cot \left(45^{\circ}\right)$

$=3 \times \frac{1}{2} \times 2-4 \times \frac{1}{2} \times 1$

$=3-2$

$=1$

= RHS

Hence proved.